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:

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:

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 = 2/5)
and a, b, c (b = 6 , c = -5) are real numbers.

Question 1(vii):

Check whether the following is quadratic equation:

Solution :

Yes.
It is an equation in the variable x of the form
ax² + bx + c = 0, where a is not equal to 0

Question 1(viii):

Check whether the following is quadratic equation:

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(ix):

Check whether the following is quadratic equation:

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:

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 = x² – 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  – 6x² – 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
10x² – 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.

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

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

Question 2(i):

Solve the quadratic equation:
x² – 196=0

Solution :

Question 2(ii):

Solve the quadratic equation:
5x² = 625

Solution :

Question 2(iii):

Solve the quadratic equation:
x² + 1 = 101

Solution :

Question 2(iv):

Solve the quadratic equation:

Solution :

>

Question 2(v):

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

Solution :

Question 2(vi):

Solve the quadratic equation:

Solution :

Question 2(vii):

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

Solution :

Question 2(viii):

Solve the quadratic equation:
-37.5x²= -37.5

Solution :

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

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

Therefore, m = 1/2 does not satisfy the equation.

Question 3(iv):

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

Solution :

Question 3(v):

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

Solution :

A number k satisfies the quadratic ax²+bx+c=0 if
ak²+bk+c=0
Given

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:

Solution :

Question 3(vii):

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

Solution :

A number k satisfies the quadratic ax²+bx+c=0 if
ak²+bk+c=0

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 :

Question 4(ii):

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

Solution :

Question 4(iii):

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

Solution :

Question 4(iv):

Solution :

Question 4(v):

Solution :

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 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:

Solution :

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:

Solution :

⟹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:

Solution :

⟹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:

Solution :

Question 15:

Solve the quadratic equation by factorization method:

Solution :

Question 16:

Solve the quadratic equation by factorization method:

Solution :

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:

Solution :

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

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

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

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

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

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

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

Question 8:

Solve the quadratic equation by completing the square:

Solution :

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

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

Exercise 9.5:

Question 1:

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

Solution :

Question 2:

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

Solution :

Question 3:

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

Solution :

Question 4:

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

Solution :

Question 5:

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

Solution :

Question 6:

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

Solution :

Question 7:

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

Solution :

Question 8:

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

Solution :

Question 9:

Solve the quadratic equation by formula method:

Solution :

Question 10:

Solve the quadratic equation by formula method:

Solution :

Question 11:

Solve the quadratic equation by formula method:

Solution :

Question 12:

Solve the quadratic equation by formula method:

Solution :

Question 13:

Solve the quadratic equation by formula method:

Solution :

Question 14:

Solve the quadratic equation by formula method:

Solution :

Exercise 9.6:

Question A(i):

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

Solution :

Question A(ii):

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

Solution :

Question A(iii):

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

Solution :

Question A(iv):

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

Solution :

Question A(v):

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

Solution :

Question A(vi):

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

Solution :

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 :

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 :

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 :

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 :

Question C(i):

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

Solution :

Question C(ii):

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

Solution :

Question C(iii):

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

Solution :

Question C(iv):

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

Solution :

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 :

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 :

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):

Solution :

Question A(iv):

Solution :

Question A(v):

Solution :

Question A(vi):

Solution :

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

Solution :

Question B(2):

If ‘a’ and ‘b’ are the roots of the equation 3m² = 6m + 5, find the value of

ii.(a + 2b)(2a + b)

Solution :

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 :

Question B(4):

Solution :

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.

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.

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)

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.

Exercise 9.9:

Question I(i):

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

Solution :

y = -x²

Question I(ii):

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

Solution :

y = 3x²

Question I(iii):

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

Solution :

y = x² + 6x

Question I(iv):

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

Solution :

y = x² – 2x

Question I(v):

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

Solution :

y= x² – 8x + 7

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

Question I(vii):

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

Solution :

y = x² + x – 6

Question I(viii):

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

Solution :

y = x² – 2x + 5

Exercise 9.10:

Question I(i):

Draw the graph of the following equation.
y = x²

Solution :

y = x²

Question I(ii):

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

Solution :

y = 3x²

Question I(iii):

Draw the graph of the following equation.
y = x²  – 4x

Solution :

y = x²  – 4x

Question I(iv):

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

Solution :

y = -x² + 8x – 16

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

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,

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

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

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

∴ 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

Solution :

Consider the figure as follows:

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
∴ x² + 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

Since side of a square cannot be negative, x = 5
∴ y² = 14 + 2x² = 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:

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

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.

∴ 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.
>
∴ 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 :

∴ 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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