KSEEB SSLC Solutions for Class 10 Maths Chapter 9 Quadratic Equations

In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 9 Quadratic Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 9 Quadratic Equations pdf, free KSEEB solutions for Class 10 Maths Chapter 9 Quadratic Equations book pdf download. Now you will get step by step solution to each question.

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations (English Medium)

KSEEB SSLC Solutions for Class 10 Maths Chapter 9 Exercise 9.1

Question 1(i):

Check whether the following is quadratic equation:
x2 – x = 0

Solution :

Yes.
It is an equation in the variable x of the form
ax2 + bx +c = 0, where a is not equal to 0 (a = 1) and
a,b ,c (b = -1, c = 0) are real numbers

Question 1(ii):

Check whether the following is quadratic equation:
x2 = 8

Solution :

x2 – 8 = 0
Yes.
It is an equation in the variable x of the form
ax2 + bx +c = 0, where a is not equal to 0 (a = 1) and a, b, c (b = 0, c = – 8) are real numbers.

Question 1(iii):

Check whether the following is quadratic equation:
x2 + ½ x = 0

Solution :

Yes.
It is an equation in the variable x of the form
ax2 + bx +c = 0, where a is not equal to 0 (a = 1) and a, b, c (b = ½, c = 0) are real numbers.

Question 1(iv):

Check whether the following is quadratic equation:
3x – 10 = 0

Solution :

No.
It is not an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 0) and a, b, c are real numbers.

Question 1(v):

Check whether the following is quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(v)

Solution :

Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 1) and a, b, c (b = – 29/4 , c = 5) are real numbers.

Question 1(vi):

Check whether the following is quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(vi)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(vi)s
Yes.
It is an equation in the variable x of the form
ax2 + bx +c = 0, where a is not equal to 0 (a = 2/5)
and a, b, c (b = 6 , c = -5) are real numbers.

Question 1(vii):

Check whether the following is quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(vii)

Solution :

Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(vii)s

Question 1(viii):

Check whether the following is quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(viii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(viii)s
No.
It is not an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 0) and a, b ,c are real numbers.

Question 1(ix):

Check whether the following is quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(ix)s

Solution :

No.
It is not an equation in the variable x of the form
ax2 + bx + c = 0, x is raised to the power 3

Question 1(x):

Check whether the following is quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-1(x)

Solution :

Yes .It is a quadratic in both two variables, x and y.

Question 2(i):

Simplify and check whether the following is quadratic equation:
x(x + 6) = 0

Solution :

x(x + 6) = 0
x2 + 6x = 0
Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 1)
and a, b, c (b = 6 , c = 0) are real numbers.

Question 2(ii):

Simplify and check whether the following is quadratic equation:
(x – 4)(2x – 3) = 0

Solution :

(x – 4)(2x – 3) = 0
2x2  – 11x + 12 = 0
Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 2)
and a, b ,c (b = – 11 , c = 12 are real numbers.

Question 2(iii):

Simplify and check whether the following is quadratic equation:
(x + 9)(x – 9) = 0

Solution :

(x + 9)(x – 9) = 0
x2 – 81 = 0
Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 1)
and a, b, c (b = 0 , c= – 81) are real numbers.

Question 2(iv):

Simplify and check whether the following is quadratic equation:
(x + 2)(x – 7) = 5

Solution :

( x + 2)(x – 7) = 5
⟹x2 – 5x – 14 – 5= 0
⟹x2 – 5x – 19= 0
Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 1)
and a, b, c (b = -5 , c = -19) are real numbers.

Question 2(v):

Simplify and check whether the following is quadratic equation:
3x + (2x – 1)(x – 9) = 0

Solution :

3x + 2x2 – 19x + 9 = 0
2x2 – 16x + 9 = 0
Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 2)
and a, b, c (b = -16 , c = 9) are real numbers.

Question 2(vi):

Simplify and check whether the following is quadratic equation:
(x + 1)2 = 2(x – 3)

Solution :

x2 + 2x + 1 = 2x – 6
x2 + 7 = 0
Yes
It is an equation in the variable x of the form
ax2 + bx +c = 0, where a is not equal to 0 (a = 1)
and a, b, c (b = 0 , c = 0) are real numbers.

Question 2(vii):

Simplify and check whether the following is quadratic equation:
(2x – 1)(x – 3) = (x + 5)(x – 1)

Solution :

2x2 – 7x + 3 = x2 + 4x – 5
x2 – 11x + 8 = 0
Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 1)
and a, b, c (b = – 11 , c = 8) are real numbers.

Question 2(viii):

Simplify and check whether the following is quadratic equation:
x2  + 3x + 1 = (x – 2)2

Solution :

x2  + 3x + 1 = (x – 2)2
x2 + 3x +1 = x– 4x + 4
7x – 3 = 0
No.
It is not an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 0)
and a, b, c are real numbers.

Question 2(ix):

Simplify and check whether the following is quadratic equation:
(x + 2)3 = 2x(x2 – 1)

Solution :

(x + 2)3 = 2x(x2 – 1)
x3 + 3(2x)(x + 2) + 8 = 2x3 – 2x
x3 + 6x2 + 12x + 8 = 2x3 – 2x
x3  – 6x– 14x – 8 = 0
No.
It is an equation in the variable x, where x is raised to the power 3. For quadratic equations the highest power of x must be 2.

Question 2(x):

Simplify and check whether the following is quadratic equation:
x3 – 4x2 – x + 1 = (x – 2)3

Solution :

x3 – 4x2 – x + 1 = (x – 2)3
x3 – 4x2 – x + 1 = x3 + 3(2x)(x – 2) – 8
x3 – 4x2 – x + 1 = x3 + 6x2 – 12x – 8
10x– 11x – 9 = 0
Yes.
It is an equation in the variable x of the form
ax2 + bx + c = 0, where a is not equal to 0 (a = 10)
and a, b, c (b= – 11, c= -9) are real numbers.

Question 3(i):

Represent in the form of quadratic equations:
The product of two consecutive integers is 306

Solution :

Let x be the integer
Next consecutive integer will be x+1
Given : x(x + 1) = 306
Therefore,x2 + x – 306 = 0

Question 3(ii):

Represent in the form of quadratic equations:
The length of a rectangular park (in metres) is one more than twice its breadth and its area is 528 m2.

Solution :

Let x be the breadth of the park
Now length of the park will be 2x + 1
Area of rectangle = length x breadth
Given : x(2x+1) = 528
Therefore,2x2 + x – 528 = 0

Question 3(iii):

Represent in the form of quadratic equations:
A train travels a distance of 480 km at uniform speed. If the speed had been 8 km/hr less, then it would have taken three more hours to cover the same distance.
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.1-3(iii)

Solution :

Let uniform speed be x
Time taken to cover 480 km = 480/x
Distance = speed × time
Given : speed = (x – 8) km/hr
Time = (480/x) + 3
Distance = 480

KSEEB SSLC Solutions for Class 10 Maths Chapter 9 Exercise 9.2

Question 1:

Classify the following equations into pure and adfected quadratic equations:

i. x2 = 100
ii. x2 + 6 = 6
iii. p(p – 3) = 1
iv. x2 +3 = 2x
v. (x + 9)(x – 9)
vi. 2x2 = 72
vii. x2 – x = 0
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-1

Solution :

Quadratic equations having variables in second degree only are called pure quadratic equations.
Quadratic equations having variables both in the first and second degree are called adfected quadratic equations.
i. x2 = 100
Only variable in second degree is present
Pure
ii. x2 + 6 = 6
x2 = 0
Only variable in second degree is presentPure
iii. p(p – 3)=1
p2 – 3p – 1=0
Variable in both first and second degree is present
Adfected
iv. x2 +3=2x
x2 – 2x + 3 =0
Variable in both first and second degree is present
Adfected
v. (x + 9)(x – 9)
x2 – 81=0
Only variable in second degree is present
Pure
vi. 2x2 = 72
Only variable in second degree is present.
Pure
vii. x2 – x =0
Variable in both first and second degree is present
Adfected
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-1s

Question 2(i):

Solve the quadratic equation:
x2 – 196=0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(i)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(i)s

Question 2(ii):

Solve the quadratic equation:
5x2 = 625

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(ii)s

Question 2(iii):

Solve the quadratic equation:
x2 + 1 = 101

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(iii)s

Question 2(iv):

Solve the quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(iv)

Solution :

>KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(iv)s

Question 2(v):

Solve the quadratic equation:
(x + 8)2 – 5 = 31

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(v)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(v)s

Question 2(vi):

Solve the quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(vi)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(vi)s

Question 2(vii):

Solve the quadratic equation:
-4x2 + 324=0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(vii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(vii)s

Question 2(viii):

Solve the quadratic equation:
-37.5x2= -37.5

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(viii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-2(viii)s

Question 3(i):

Determine whether the given value of x satisfies the quadratic equation:
x2 +14x+13=0; x= – 1,x= – 13

Solution :

A number k satisfies the quadratic ax2+bx+c=0 if
ak2+bk+c=0
Given x2 +14x+13=0; x=-1,x=-13
Checking for x= -1
(-1)2+14(-1)+13= 1-14+13=0
Therefore, x=-1 satisfies the equation
Checking for x= -13
(-13)2+14(-13)+13=169-182+13=0
Therefore, x=-13 satisfies the equation

Question 3(ii):

Determine whether the given value of x satisfies the quadratic equation:
7x2 – 12x = 0; x=1/3

Solution :

A number k satisfies the quadratic ax2+bx+c=0 if
ak2+bk+c = 0
Given 7x2 – 12x = 0; x = 1/3
Checking for x= 1/3
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(ii)s
Therefore, x=1/3 does not satisfy the equation

Question 3(iii):

Determine whether the given value of x satisfies the quadratic equation:
2m2 – 6m + 3 = 0; m = 1/2

Solution :

A number k satisfies the quadratic ax2+bx+c=0 if
ak2+bk+c = 0
Given 2m2 – 6m + 3 = 0; m = 1/2
Checking for m = 1/2
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(iii)s
Therefore, m = 1/2 does not satisfy the equation.

Question 3(iv):

Determine whether the given value of x satisfies the quadratic equation:

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(iv)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(iv)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(iv)s

Question 3(v):

Determine whether the given value of x satisfies the quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(v)

Solution :

A number k satisfies the quadratic ax2+bx+c=0 if
ak2+bk+c=0
Given
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(v)s
Simplifying the equation we get
3x2 -2x-1=0
Checking for x= 1
3(1)2 – 2(1) – 1 = 3 – 2 – 1 = 0
Therefore, x = 1 satisfies the equation
Checking for x = – 1
3( – 1)2 – 2(- 1) – 1 = 3 + 2 – 1 = 4
Therefore, x = – 1 does not satisfy the equation

Question 3(vi):

Determine whether the given value of x satisfies the quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(vi)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(vi)s

Question 3(vii):

Determine whether the given value of x satisfies the quadratic equation:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(vii)

Solution :

A number k satisfies the quadratic ax2+bx+c=0 if
ak2+bk+c=0
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-3(vii)s
Simplifying we get
2x = x + 2
Therefore, x=2

Question 3(viii):

Determine whether the given value of x satisfies the quadratic equation:
6x2 -x – 2 = 0; x = -1/2, x = 2/3

Solution :

A number k satisfies the quadratic ax2+bx+c=0 if
ak2 + bk + c = 0
Given 6x2 – x – 2 = 0; x = -1/2, x = 2/3
Checking for x = – 1/2
6(-1/2)2 – (-1/2) – 2 = 3/2 + 1/2 – 2 = 0
Therefore, x = -1/2 satisfies the equation
Checking for x = 2/3
6(2/3)2 – (2/3) – 2 = 8/3 – 2/3 – 2 = 0
Therefore, x = 2/3 satisfies the equation.

Question 4(i):

If A = πr2 solve for r, and find the value of r if A = 77 and π = 22/7.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(i)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(i)s

Question 4(ii):

If r2= l2 + d2, solve for d, and find the value of d if r = 5 and l = 4.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(ii)s

Question 4(iii):

If c2= a2+b2, solve for b, and find the value of b if a = 8 and c = 17.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(iii)s

Question 4(iv):

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(iv)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(iv)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(iv)s

Question 4(v):

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(v)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(v)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(v)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(v)s

Question 4(vi):

If v2 = u2 + 2as, solve for v, and find the value of v if u = 0, a = 2 and s = 100.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.2-4(vi)s

KSEEB SSLC Solutions for Class 10 Maths Chapter 9 Exercise 9.3

Question 1:

Solve the quadratic equation by factorization method:
x2  + 15x + 50 = 0

Solution :

x2 + 15x + 50 = 0
⟹ x2 + 5x + 10x + 50 = 0
⟹ x(x + 5) + 10(x + 5) = 0
⟹ (x + 10)(x + 5) = 0
⟹ x + 10 = 0 or x+5 = 0
⟹ x = – 10 or x = -5

Question 2:

Solve the quadratic equation by factorization method:
x2 – 3x – 10 = 0

Solution :

x2 – 3x – 10 = 0
⟹ x2 + 2x – 5x – 10 = 0
⟹ (x + 2) – 5(x + 2) = 0
⟹ (x + 2)(x – 5) = 0
⟹ x + 2 = 0 or x – 5 = 0
⟹ x = -2 or x = 5

Question 3:

Solve the quadratic equation by factorization method:
6 – p2 = p

Solution :

6 – p2 = p
⟹ p2 + p – 6 = 0
⟹ p2 + 3p – 2p – 6 = 0
⟹p(p + 3) – 2(p + 3) = 0
⟹(p – 2)(p + 3) = 0
⟹p – 2 = 0 or p + 3 = 0
⟹p = 2 or p = -3

Question 4:

Solve the quadratic equation by factorization method:
2x2 + 5x – 12 = 0

Solution :

2x2 + 5x – 12 = 0
⟹2x2 + 5x – 12 = 0
⟹2x2 + 8x – 3x – 12 = 0
⟹2x(x + 4) – 3(x + 4) = 0
⟹(2x – 3)(x + 4) = 0
⟹(2x – 3) = 0 or (x + 4) = 0
⟹x = 3/2 or x = -4

Question 5:

Solve the quadratic equation by factorization method:
13m = 6(m2 + 1)

Solution :

13m = 6(m2 + 1)
⟹6m2 – 13m + 6 = 0
⟹6m2 – 9m – 4m + 6 = 0
⟹3m(2m – 3) – 2(2m – 3) = 0
⟹(3m – 2)(2m – 3) = 0
⟹3m – 2 = 0 or 2m – 3 = 0
⟹m = 2/3 or m = 3/2

Question 6:

Solve the quadratic equation by factorization method:
100x2 – 20x + 1 = 0

Solution :

100x2 – 20x + 1= 0
⟹100x2 – 10x – 10x + 1= 0
⟹10x(10x – 1) – 1(10x – 1) = 0
⟹(10x – 1)(10x – 1) = 0
⟹(10x – 1)2= 0
x = 1/10 is a repeated root of the equation

Question 7:

Solve the quadratic equation by factorization method:

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-7
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-7

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-7s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-7s

Question 8:

Solve the quadratic equation by factorization method:
x2 + 4kx + 4k2 = 0

Solution :

x2 + 4kx + 4k2 = 0
⟹x2 + 2kx + 2kx + 4k2 = 0
⟹x(x + 2k) + 2k(x + 2k) = 0
⟹(x + 2k)(x + 2k) = 0
⟹(x + 2k)2 =0
x = -2k is a repeated root of the equation

Question 9:

Solve the quadratic equation by factorization method:

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-9
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-9

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-9s
⟹m2 – 6m – 7 = 0
⟹m2 – 7m + m – 7 = 0
⟹m(m – 7) + 1(m – 7) = 0
⟹(m + 1)(m – 7) = 0
⟹m + 1= 0 or m-7=0
⟹m = -1 or m = 7

Question 10:

Solve the quadratic equation by factorization method:

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-10
Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-10
⟹2x2 – 5x + 2 = 0
⟹2x2 – 4x – x + 2 = 0
⟹2x(x – 2) – 1(x – 2) = 0
⟹(2x – 1)(x – 2) = 0
⟹2x – 1 = 0 or x – 2 = 0
⟹x = 1/2 or x = 2

Question 11:

Solve the quadratic equation by factorization method:
21y2 = 62y + 3

Solution :

21y2 = 62y + 3
⟹21y2 – 62y – 3 = 0
⟹21y2 – 63y + y-3 = 0
⟹21y(y – 3) + 1(y – 3) = 0
⟹(21y + 1)(y – 3) = 0
⟹21y + 1= 0 or y – 3 = 0
⟹y = -1/21 or y = 3

Question 12:

Solve the quadratic equation by factorization method:
0.2t2  – 0.04t = 0.03

Solution :

0.2t2 – 0.04t = 0.03
⟹0.2t2  – 0.04t – 0.03 = 0
⟹Multiplying by 100 on both sides
⟹20t2 – 4t – 3 = 0
⟹20t2 – 10t + 6t – 3 = 0
⟹10t(2t – 1) + 3(2t – 1)=0
⟹10t + 3 = 0 or 2t – 1= 0
⟹t = -3/10 or t = 1/2
Note: Question is wrong. Middle term will be 0.04 instead of 0.04.

Question 13:

Solve the quadratic equation by factorization method:
4x2 + 32x + 64 = 0

Solution :

4x2 + 32x + 64 = 0
Divide by 4 throughout the equation
⟹x2+ 8x +16 = 0
⟹(x + 4)2= 0
⟹x + 4 = 0
⟹x = – 4 is a repeated root of the equation

Question 14:

Solve the quadratic equation by factorization method:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-14

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-14s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-14s

Question 15:

Solve the quadratic equation by factorization method:

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-15
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-15

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-15s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-15s

Question 16:

Solve the quadratic equation by factorization method:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-16

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-16s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-16s

Question 17:

Solve the quadratic equation by factorization method:
a2b2x2 – (a2 + b2)x + 1 = 0

Solution :

a2b2x2 – (a2 + b2)x + 1 = 0
⟹a2b2x2 – a2x – b2x + 1= 0
⟹a2x(b2x – 1) – 1(b2x – 1) = 0
⟹(b2x – 1)(a2x – 1) = 0
⟹(b2x – 1) = 0 or (a2x – 1) = 0
⟹x = 1/b2 or x = 1/a2

Question 18:

Solve the quadratic equation by factorization method:
2(x + 1)2 – 5(x + 1) = 12

Solution :

2(x + 1)2 – 5(x + 1) = 12
⟹2(x2 + 2x + 1) – 5x – 5 – 12 = 0
⟹2x2 – x – 15 = 0
⟹2x2 – 6x + 5x – 15 = 0
⟹2x(x – 3) + 5(x – 3) = 0
⟹(2x + 5)(x – 3) = 0
⟹x = -5/2 or x = 3

Question 19:.

Solve the quadratic equation by factorization method:
(x – 4)2+122 = 152

Solution :

(x – 4)2 + 122 = 152
⟹x2 – 8x + 16 + 144 = 225
⟹x2 – 8x + 65 = 0
⟹x2 – 13x + 5x – 65 = 0
⟹x(x – 13) + 5(x – 13) = 0
⟹(x + 5)(x – 13) = 0
⟹x = – 5 or x = 13

Question 20:

Solve the quadratic equation by factorization method:

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-20
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-20

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-20s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.3-20s

Exercise 9.4:

Question 1:

Solve the quadratic equation by completing the square:
4x2 – 20x + 9 = 0

Solution :

4x2 – 20x + 9 = 0
Dividing by coefficient of x2 on both sides
⟹ x2 – 5x + 9/4 = 0
Adding and subtracting the square of half the coefficient of x
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-1s

Question 2:

Solve the quadratic equation by completing the square:
4x2 + x – 5 = 0

Solution :

4x2 + x – 5 = 0
Dividing by coefficient of x2 on both sides
⟹ x2 +1/4x – 5/4 = 0
Adding and subtracting the square of half the coefficient of x
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-2s

Question 3:

Solve the quadratic equation by completing the square:
2x2 + 5x – 3 = 0

Solution :

2x2 + 5x – 3 = 0
Dividing by coefficient of x2 on both sides
⟹ x2 + 5/2 x – 3/2 = 0
Adding and subtracting the square of half the coefficient of x
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-3s

Question 4:

Solve the quadratic equation by completing the square:
x2 + 16x – 9 = 0

Solution :

x2  + 16x – 9 = 0
Adding and subtracting the square of half the coefficient of x
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-4s

Question 5:

Solve the quadratic equation by completing the square:
x2 – 3x + 1= 0

Solution :

x2 – 3x + 1 = 0
Adding and subtracting the square of half the coefficient of x
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-5s

Question 6:

Solve the quadratic equation by completing the square:
t2  + 3t = 7

Solution :

t2 +3t – 7 = 0
Adding and subtracting the square of half the coefficient of t
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-6s

Question 7:

Solve the quadratic equation by completing the square:
3x(x – 5) = 2x(x + 7)

Solution :

3x(x – 5) = 2x(x + 7)
⟹ 3x2 – 15x = 2x2 + 14x
⟹ x2 – 29x = 0
Adding and subtracting the square of half the coefficient of x
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-7s

Question 8:

Solve the quadratic equation by completing the square:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-8

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-8s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-8s

Question 9:

Solve the quadratic equation by completing the square:
a2x2 – 3abx + 2b2 = 0

Solution :

a2x2  – 3abx + 2b2
Dividing by coefficient of x2 on both sides
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-9s

Question 10:

Solve the quadratic equation by completing the square:
4x2 + 4bx – (a2 – b2) = 0

Solution :

4x2 + 4bx – (a2 – b2) = 0
Dividing by coefficient of x2 on both sides
⟹ x2 + bx – (a2 – b2)/4 = 0
Adding and subtracting the square of half the coefficient of x
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.4-10s

Exercise 9.5:

Question 1:

Solve the quadratic equation by formula method:
x2 – 4x + 2=0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-1s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-1s

Question 2:

Solve the quadratic equation by formula method:
x2 – 2x + 4 = 0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-2s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-2s

Question 3:

Solve the quadratic equation by formula method:
x2 – 7x + 12 = 0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-3s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-3s

Question 4:

Solve the quadratic equation by formula method:
2y2 + 6y = 3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-4s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-4s

Question 5:

Solve the quadratic equation by formula method:
5m2 – 11m + 2 = 0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-5s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-5s

Question 6:

Solve the quadratic equation by formula method:
8r2 = r + 2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-6s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-6s

Question 7:

Solve the quadratic equation by formula method:
p = 5 – 2p2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-7s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-7s

Question 8:

Solve the quadratic equation by formula method:
(2x + 3)(3x – 2) + 2 = 0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-8s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-8s

Question 9:

Solve the quadratic equation by formula method:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-9

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-9s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-9s

Question 10:

Solve the quadratic equation by formula method:

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-10
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-10

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-10s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-10s

Question 11:

Solve the quadratic equation by formula method:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-11

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-11s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-11s

Question 12:

Solve the quadratic equation by formula method:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-12

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-12
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-12

Question 13:

Solve the quadratic equation by formula method:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-13

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-13s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-13s

Question 14:

Solve the quadratic equation by formula method:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-14

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-14s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.5-14s

Exercise 9.6:

Question A(i):

Discuss the nature of the roots of the equation:
y2  – 7y + 2 = 0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(i)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(i)

Question A(ii):

Discuss the nature of the roots of the equation:
x2 -2x+3=0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(i)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(i)

Question A(iii):

Discuss the nature of the roots of the equation:
2n2 +5n-1=0c

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(iii)s

Question A(iv):

Discuss the nature of the roots of the equation:
a2 +4a+4=0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(iv)s

Question A(v):

Discuss the nature of the roots of the equation:
x2 +3x-4=0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(v)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(v)s

Question A(vi):

Discuss the nature of the roots of the equation:
3d2 – 2d + 1= 0

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(vi)s

Question B(i):

For what positive values of m are the roots of the equation a2 – ma + 1,
1.Equal 2.Distinct 3.Imaginary

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-A(vi)s

Question B(ii):

For what positive values of m are the roots of the equation x2-mx+9,
1.Equal 2.Distinct 3.Imaginary

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(ii)s

Question B(iii):

For what positive values of m are the roots of the equation r2 – (m + 1)r + 4,
1.Equal 2.Distinct 3.Imaginary

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iii)s

Question B(iv):

For what positive values of m are the roots of the equation mk2-3k+1,
1.Equal 2.Distinct 3.Imaginary

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iv)s

Question C(i):

Find the values of p for which roots of the equation
x2 – px + 9 = 0 are equal.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iv)s

Question C(ii):

Find the values of p for which roots of the equation
2a2+3a+p =0 are equal

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-B(iv)s

Question C(iii):

Find the values of p for which roots of the equation
pk2-12k+9=0 are equal

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(iii)s

Question C(iv):

Find the values of p for which roots of the equation
2y2-py+1=0 are equal

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(iv)s

Question C(v):

Find the values of p for which roots of the equation
(p+1)n2+2(p+3)n+(p+8)=0 are equal

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(v)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(v)s

Question C(vi):

Find the values of p for which roots of the equation
(3p+1)c2+2(p+1)c + p = 0 are equal

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.6-C(vi)s

Exercise 9.7:

Question 1:

Find the sum and product of the roots of the following quadratic equation:
x2 -5x+8=0

Solution :

This is in the form of ax2+bx+c=0
a = 1,b =-5,c = 8
Sum of roots = -b/a = 5/1= 5
Product of roots = c/a = 8/1 = 8

Question 2:

Find the sum and product of the roots of the following quadratic equation:
3a2  – 10a – 5 = 0

Solution :

This is in the form of pa2 + qa + r = 0
p = 3,q = -10, r =-5
Sum of roots = -q/p = 10/3
Product of roots = r/p = -5/3

Question 3:

Find the sum and product of the roots of the following quadratic equation:
8m2 -m=2

Solution :

8m2 -m=2
8m2 – m – 2 = 0
This is in the form of am2 + bm + c =0
a = 8, b = -1, c =-2
Sum of roots = -b/a = 1/8
Product of roots = c/a = -2/8 = -1/4

Question 4:3

Find the sum and product of the roots of the following quadratic equation:
6k2 – 3 = 0

Solution :

This is in the form of ak2+bk+c=0
a = 6, b = 0, c =-3
Sum of roots = -b/a = 0/6 = 0
Product of roots = c/a = -3/6 = -1/2

Question 5:

Find the sum and product of the roots of the following quadratic equation:
pr2 – r – 5 = 0

Solution :

This is in the form of ar2+br+c=0
a = p, b =-1, c = -5
Sum of roots = -b/a = 1/p
Product of roots = c/a = -5/p

Question 6:

Find the sum and product of the roots of the following quadratic equation:
x2 + (ab)x + (a+b)=0

Solution :

This is in the form of px2 + qx + r =0
p =1,q = ab , r = a+b
Sum of roots = -q/p = -ab/1 = -ab
Product of roots = r/p = (a+b)/1 = a+b

Exercise 9.8:

Question A(i):

Form the equation whose roots are 3, 5.

Solution :

Let m and n be the roots.
∴ m = 3 and n = 5

Sum of the roots = m + n = 3 + 5 = 8
Product of the roots = mn = 3 × 5 = 15

Standard form is x2 – (m + n)x + mn = 0
∴ The required quadratic equation is x2 – 8x + 15 = 0.

Question A(ii):

Form the equation whose roots are 6, -5.

Solution :

Let m and n be the roots.
∴ m = 6 and n = -5

Sum of the roots = m + n = 6 + (-5) = 1
Product of the roots = mn = 6 × (-5) = -30

Standard form is x2 – (m + n)x + mn = 0
∴ The required quadratic equation is x2 – x – 30 = 0.

Question A(iii):

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iii)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iii)s

Question A(iv):

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iv)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iv)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(iv)s

Question A(v):

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(v)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(v)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(v)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(v)s

Question A(vi):

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(vi)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(vi)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-A(vi)s

Question B(1):

If ‘m’ and ‘n’ are the roots of the equation x2 – 6x + 2 = 0, find the value of
i. (m + n)mn
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(1)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(1)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(1)s

Question B(2):

If ‘a’ and ‘b’ are the roots of the equation 3m2 = 6m + 5, find the value of
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(2)
ii.(a + 2b)(2a + b)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(2)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(2)s

Question B(3):

If ‘p’ and ‘q’ are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of
i. (p + q)2+ 4pq
ii. p3+ q3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(3)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(3)s

Question B(4):

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(4)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(4)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(4)s
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(4)s

Question B(5):

Find the value of ‘k’ so that the equation x2 + 4x + (k + 2) = 0 has one root equal to zero.

Solution :

Given equation is x2 + 4x + (k + 2) = 0.
∴ a = 1, b = 4, c = (k + 2)
Let one root of the equation be ‘m’ and other root is zero.
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(5)s

Question B(6):

Find the value of ‘q’ so that the equation 2x2 – 3qx + 5q = 0 has one root which is twice the other.

Solution :

Given equation is 2x2 – 3qx + 5q = 0
∴ a = 2, b = -3q and c = 5q
Let one root of the equation be ‘m’.
Then the other root will be 2m.
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(5)s

Question B(7):

Find the value of ‘p’ so that the equation 4x2 – 8px + 9 = 0 has roots whose difference is 4.

Solution :

Given equation is 4x2 – 8px + 9 = 0
∴ a = 4, b = -8p, c = 9
Let one root of the equation be ‘m’.
Then, the other root will be (m – 4)
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(5)s

Question B(8):

If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.

Solution :

Given equation is x2 + px + q = 0
∴ a = 1, b = p, c = q
Let one root of the equation be ‘m’.
Then, the other root will be 3m.
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.8-B(8)s

Exercise 9.9:

Question I(i):

Draw the graph of the following quadratic equation:
y = -x2

Solution :

y = -x2

x-3-2-10123
y-9-4-10-1-4-9

Question I(ii):

Draw the graph of the following quadratic equation:
y = 3x2

Solution :

y = 3x2

x-3-2-10123
y27123031227

Question I(iii):

Draw the graph of the following quadratic equation:
y = x+ 6x

Solution :

y = x+ 6x

x-7-6-5-4-3-2-101
y70-5-8-9-8-507

Question I(iv):

Draw the graph of the following quadratic equation:
y = x2 – 2x

Solution :

y = x2 – 2x

x-2-101234
y830-1038

Question I(v):

Draw the graph of the following quadratic equation:
y= x2 – 8x + 7

Solution :

y= x2 – 8x + 7

x-1012478
y1670-5-907

Question I(vi):

Draw the graph of the following quadratic equation:
y = (x + 2)(2 – x)

Solution :

y = (x + 2)(2 – x)
∴ y = 2x – x2 + 4 – 2x
∴ y = -x2 + 4

x-3-2-10123
y-503430-5

Question I(vii):

Draw the graph of the following quadratic equation:
y = x2 + x – 6

Solution :

y = x2 + x – 6

x-4-3-2-10123
y60-4-6-6-406

Question I(viii):

Draw the graph of the following quadratic equation:
y = x2 – 2x + 5

Solution :

y = x2 – 2x + 5

x-2-101234
y138545813

Exercise 9.10:

Question I(i):

Draw the graph of the following equation.
y = x2

Solution :

y = x2

x-3-2-10123
y9410149

Question I(ii):

Draw the graph of the following equation.
y = 3x2

Solution :

y = 3x2

x-2-1012
y1230312

Question I(iii):

Draw the graph of the following equation.
y = x – 4x

Solution :

y = x – 4x

x-1012345
y50-3-4-305

Question I(iv):

Draw the graph of the following equation.
y = -x2 + 8x – 16

Solution :

y = -x2 + 8x – 16

x1234567
y-9-4-10-1-4-9

Exercise 9.11:

Question 1:

Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.

Solution :

Let the two consecutive positive odd numbers be x and (x + 2).
∴ x2 + (x + 2)2 = 130
∴ x2 + x2 + 4x + 4 = 130
∴ 2x2 + 4x + 4 – 130 = 0
∴ 2x2 + 4x – 126 = 0
Dividing by 2 throughout, we get
x2 + 2x – 63 = 0
∴ x2 + 9x – 7x – 63 = 0
∴ x(x + 9) – 7(x + 9) = 0
∴ (x + 9)(x – 7) = 0
∴ x = -9 or x = 7
But x is positive odd number.
∴ x = 7 and x + 2 = 7 + 2 = 9
Thus, the two consecutive positive odd numbers are 7 and 9.

Question 2:

Find the whole number such that four times the number subtracted from three times the square of the number makes 15.

Solution :

Let the whole number be x.
∴ 3x2 – 4x = 15
∴ 3x2 – 4x – 15 = 0
∴ 3x2 – 9x + 5x – 15 = 0
∴ 3x(x – 3) + 5(x – 3) = 0
∴ (x – 3)(3x + 5) = 0
∴ 3x + 5 = 0 or x – 3 = 0
∴ 3x = -5 or x = 3
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-2s
But x is a whole number.
∴ x = 3
Thus, the whole number is 3.

Question 3:

The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is 8/15.

Solution :

Let the two natural numbers be x and y.
∴ x + y = 8
∴ x = 8 – y ….(1)
Also,
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-3s

Question 4:

A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number.

Solution :

Let the digit in the unit place be x and the digit in the tenth place by y.
Thus, the number formed is 10y + x.
Now, x × y = 12 ….(1)
Again, the number formed by interchanging the digits is 10x + y.
According to given information, we have
10y + x + 36 = 10x + y
∴ 10x + y – x – 10y = 36
∴ 9x – 9y = 36
Dividing by 9 throughout, we get
x – y = 4
∴ x = 4 + y ….(2)
Substituting (2) in (1), we get
(4 + y) × y = 12
∴ 4y + y2 = 12
∴ y2 + 4y – 12 = 0
∴ y2 + 6y – 2y – 12 = 0
∴ y(y + 6) – 2(y + 6) = 0
∴ (y + 6)(y – 2) = 0
∴ y + 6 = 0 or y – 2 = 0
∴ y = -6 or y = 2
Since digit cannot be negative, y = 2
∴ x = 4 + y = 4 + 2 = 6
Thus, the number is 10y + x = 10(2) + 6 = 20 + 6 = 26.

Question 5:

Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.

Solution :

Let the three consecutive positive integers be x, (x + 1) and (x + 2).
∴ x2 + (x + 1)(x + 2) = 154
∴ x2 + x2 + 2x + x + 2 – 154 = 0
∴ 2x2 + 3x – 152 = 0
∴ 2x2 + 19x – 16x – 152 = 0
∴ x(2x + 19) – 8(2x + 19) = 0
∴ (2x + 19)(x – 8) = 0
∴ 2x + 19 = 0 or x – 8 = 0
∴ 2x = -19 or x = 8
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-5s
Since x is a positive integer, x = 8
∴ x + 1 = 8 + 1 = 9
x + 2 = 8 + 2 = 10
Thus, the three consecutive positive integers are 8, 9 and 10.

Question 6:

The ages of Kavya and Karthik are 11 years and 14 years. In how many years time will the product of their ages be 304.

Solution :

Let the product of the ages of Kavya and Karthik is 304 in x years.
Then, after x years,
Kavya’s age = (11 + x) years
Karthik’s age = (14 + x) years
∴ (11 + x)(14 + x) = 304
∴ 154 + 11x + 14x + x2 – 304 = 0
∴ x2 + 25x – 150 = 0
∴ x2 + 30x – 5x – 150 = 0
∴ x(x + 30) – 5(x + 30) = 0
∴ (x + 30)(x – 5) = 0
∴ x + 30 = 0 or x – 5 = 0
∴ x = -30 or x = 5
Since a year cannot be negative, x = 5.
Thus, in 5 years time, the product of their ages will be 304.

Question 7:

The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find their present age.

Solution :

Let the present age of son be x years.
Then, the present age of his father is (2x2) years.
After eight years,
Son’s age = (x + 8) years
Father’s age = (2x2 + 8) years
According to given information, we have
(2x2 + 8) = 3(x + 8) + 4
∴ 2x2 + 8 = 3x + 24 + 4
∴ 2x2 – 3x + 8 – 28 = 0
∴ 2x2 – 3x – 20 = 0
∴ 2x2 – 8x + 5x – 20 = 0
∴ 2x(x – 4) + 5(x – 4) = 0
∴ (x – 4)(2x + 5) = 0
∴ x – 4 = 0 or 2x + 5 = 0
∴ x = 4 or 2x = -5
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-7s
Since, age cannot be negative, x = 4
∴ 2x2 = 2(4)2 = 2(16) = 32
Thus, present age of the son is 4 years and that of his father is 32 years.

Question 8:

The area of a rectangle is 56 cm. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.

Solution :

Base of a rectangle = Length of a rectangle = x + 5
Height of a rectangle = Breadth of a rectangle = x – 5
Area of a rectangle = 56 cm
Now, area of a rectangle = Length × Breadth
∴ 56 = (x + 5)(x – 5)
∴ 56 = x2 – (5)2
∴ 56 = x2 – 25
∴ x2 = 56 + 25
∴ x2 = 81
∴ x =  ±9
Since the dimensions of a rectangle cannot be negative, x = 9.
∴ Base of a rectangle = x + 5 = 9 + 5 = 14 cm
Height of a rectangle = x – 5 = 9 – 5 = 4 cm
Thus, the dimensions of a rectangle are 14 cm and 4 cm.

Question 9:

The altitude of a triangle is 6 cm greater than its base. If its area is 108 cm2. Find its base and height.

Solution :

Let the base of the triangle be ‘x’ cm.
Then, altitude (height) of a triangle = (x + 6) cm
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-9s
∴ 216 = x2 + 6x
∴ x2 + 6x – 216 = 0
∴ x2 + 18x – 12x – 216 = 0
∴ x(x + 18) – 12(x + 18) = 0
∴ (x + 18)(x – 12) = 0
∴ x = -18 or x = 12
Since the base of a triangle cannot be negative, x = 12.
∴ Height = x + 6 = 12 + 6 = 18
Thus, the base of a triangle is 12 cm and its height is 18 cm.

Question 10:

In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = x + 7, and AB = x + 8, find the lengths of the diagonals
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-10

Solution :

Consider the figure as follows:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-10s
Diagonals AC and BD bisect each other perpendicularly.
In right-angles ∆AEB, by using Pythagoras theorem, we have
AB2 = AE2 + BE2
∴ (x + 8)2 = x2 + (x + 7)2
∴ x+ 16x + 64 = x2 + x2 + 14x + 49
∴ x2 + 14x + 49 – 16x – 64 = 0
∴ x2 – 2x – 15 = 0
∴ x2 – 5x + 3x – 15 = 0
∴ x(x – 5) + 3(x – 5) = 0
∴ (x – 5)(x + 3) = 0
∴ x – 5 = 0 or x + 3 = 0
∴ x = 5 or x = -3
Since length cannot be negative, x = 5.
∴ Diagonal AC = AE + EC = x + x = 2x = 2 × 5 = 10 cm
Diagonal BD = x + 7 + x + 7 = 2x + 14 = 10 + 14 = 24 cm

Question 11:

If twice the area of smaller square is subtracted from the area of a larger square, the result is 14 cm2. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Determine the sides of the two squares.

Solution :

Let the side of the smaller square be ‘x’ cm and side of the bigger square be ‘y’ cm.
∴ Area of the smaller square = x2
 Area of the bigger square = y2
According to given information, we have
y2 – 2x2 = 14
∴ y2 = 14 + 2x2 ….(1)
Also, 2y2 + 3x2 = 203 ….(2)
Substituting (1) in (2), we have
2(14 + 2x2) + 3x2 = 203
∴ 28 + 4x2 + 3x2 – 203 = 0
∴ 7x2 – 175 = 0
∴ 7x2 = 175
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-11s
Since side of a square cannot be negative, x = 5
∴ y2 = 14 + 2x= 14 + 2(5)2 = 14 + 50 = 64
∴ y = ±8
Since side of a square cannot be negative, y = 8
Thus, the side of a smaller square is 5 cm and that of a bigger square is 8 cm.

Question 12:

In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC. If DC = x, BD = 2x – 1 and BC = 2x + 1, find the lengths of all three sides of the triangle.

Solution :.

Consider the following figure:
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-12s
In ∆ABC, BD ⊥ AC
∴ AD = DC = x
In right-angled ∆BDC, by Pythagoras theorem,
BC2 = BD2 + DC2
(2x + 1)2 = (2x – 1)2 + x2
∴ 4x2 + 4x + 1 = 4x2 – 4x + 1 + x2
∴ x2 + 4x2 – 4x2 – 4x – 4x + 1 – 1 = 0
∴ x2 – 8x = 0
∴ x(x – 8) = 0
∴ x = 0 or x = 8
Since measurement cannot be negative, x = 8.
AC = AD + DC = x + x = 2x = 2 × 8 = 16 cm
AB = BC = 2x + 1 = 2(8) + 1 = 16 + 1 = 17 cm
Thus, the lengths of sides of a triangle are 16 cm, 17 cm and 17 cm.

Question 13:

A motor boat whose speed is 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.

Solution :

Speed of motor boat in still water = 15 km/hr
Total distance travelled = 3 km
Let the speed of the stream be x km/hr.
Speed of the motor boat in downstream = (15 + x) km/hr
Speed of the motor boat in upstream = (15 – x) km/hr
Total time taken = 4 hours and 30 minutes
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-12s

Question 14:

A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.

Solution :

Selling price of the article = Rs. 24.
Let the cost price of the article be Rs. x.
KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-14s
∴ x2 = 100(24 – x)
∴ x2 = 2400 – 100x
∴ x2 + 100x – 2400 = 0
∴ x2 + 120x – 20x – 2400 = 0
∴ x(x + 120) – 20(x + 120) = 0
∴ (x + 120)(x – 20) = 0
∴ x + 120 = 0 or x – 20 = 0
∴ x = -120 or x = 20
Since, price of an article cannot be negative, x = 20.
Thus, the cost price of an article is Rs. 20.

Question 15:

Nandana takes 6 days less than the number of days taken by shobha to complete a piece of work. If both nandana and shobhatogether can complete the same work in 4 days, in how many days will shobha alone complete the work?

Solution :

Let Shobha alone can complete the work in x days.
Then, Nandana alone can complete the work in (x – 6) days.
>KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-15s
∴ 4(2x – 6) = 1(x2 – 6x)
∴ 8x – 24 = x2 – 6x
∴ x2 – 6x – 8x + 24 = 0
∴ x2 – 14x + 24 = 0
∴ x2 – 12x – 2x + 24 = 0
∴ x(x – 12) – 2(x – 12) = 0
∴ (x – 12)(x – 2) = 0
∴ x – 12 = 0 or x – 2 = 0
∴ x = 12 or x = 2
x = 2 has no meaning as 6 days less than 2 will be -4.
∴ x = 2
Thus, the number of days taken by shobha alone to complete the work is 12.

Question 16:

A particle is projected from ground level so that its height above the ground after t second is given by (20t – 5t2) m. After how many seconds is it 15 m above the ground? Can u explain briefly why there are two possible answers?

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Quadratic Equations-Ex-9.11-16s
∴ t(20 – 5t) = 15
∴ 20t – 5t2 = 15
∴ 5t2 – 20t + 15 = 0
Dividing throughout by 5, we get
t2 – 4t + 3 = 0
∴ t2 – 3t – t + 3 = 0
∴ t(t – 3) – 1(t – 3) = 0
∴ (t – 3)(t – 1) = 0
∴ t – 3 = 0 or t – 1 = 0
∴ t = 3 or t = 1
When t = 3, speed = 20 – 5t = 20 – 5 × 3 = 20 – 15 =  5 m/s
When t = 1, speed = 20 – 5t = 20 – 5 × 1 = 20 – 5 = 15 m/s

There are two answers because it reaches the height of 15 m on the way up and again on the way down.

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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