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.

Table of Contents

**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:

x^{2} – x = 0

**Solution :**

Yes.

It is an equation in the variable x of the form

ax^{2} + 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:

x^{2} = 8

**Solution :**

x^{2} – 8 = 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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:

x^{2 }+ ½ x = 0

**Solution :**

Yes.

It is an equation in the variable x of the form

ax^{2} + 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

ax^{2} + 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

ax^{2} + 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

ax^{2} + 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

ax^{2} + 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

ax^{2} + 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

ax^{2} + 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

x^{2} + 6x = 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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

2x^{2 } – 11x + 12 = 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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

x^{2 }– 81 = 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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

⟹x^{2} – 5x – 14 – 5= 0

⟹x^{2} – 5x – 19= 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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 + 2x^{2 }– 19x + 9 = 0

2x^{2 }– 16x + 9 = 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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 :**

x^{2 }+ 2x + 1 = 2x – 6

x^{2 }+ 7 = 0

Yes

It is an equation in the variable x of the form

ax^{2} + 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 :**

2x^{2 }– 7x + 3 = x^{2 }+ 4x – 5

x^{2 }– 11x + 8 = 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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:

x^{2 }+ 3x + 1 = (x – 2)^{2}

**Solution :**

x^{2 }+ 3x + 1 = (x – 2)^{2}x^{2 }+ 3x +1 = x^{2 }– 4x + 4

7x – 3 = 0

No.

It is not an equation in the variable x of the form

ax^{2} + 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(x^{2 }– 1)

**Solution :**

(x + 2)^{3} = 2x(x^{2 }– 1)

x^{3 }+ 3(2x)(x + 2) + 8 = 2x^{3 }– 2x

x^{3 }+ 6x^{2 }+ 12x + 8 = 2x^{3 }– 2x

x^{3 }– 6x^{2 }– 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:

x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}

**Solution :**

x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}x^{3} – 4x^{2} – x + 1 = x^{3 }+ 3(2x)(x – 2) – 8

x^{3} – 4x^{2} – x + 1 = x^{3 }+ 6x^{2 }– 12x – 8

10x^{2 }– 11x – 9 = 0

Yes.

It is an equation in the variable x of the form

ax^{2} + 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,x^{2 }+ 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 m^{2}.

**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,2x^{2 }+ 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. x^{2 }= 100

ii. x^{2 }+ 6 = 6

iii. p(p – 3) = 1

iv. x^{2 }+3 = 2x

v. (x + 9)(x – 9)

vi. 2x^{2} = 72

vii. x^{2 }– 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. x^{2 }= 100

Only variable in second degree is present

Pure

ii. x^{2 }+ 6 = 6

x^{2 }= 0

Only variable in second degree is presentPure

iii. p(p – 3)=1

p^{2 }– 3p – 1=0

Variable in both first and second degree is present

Adfected

iv. x^{2 }+3=2x

x^{2 }– 2x + 3 =0

Variable in both first and second degree is present

Adfected

v. (x + 9)(x – 9)

x^{2 }– 81=0

Only variable in second degree is present

Pure

vi. 2x^{2} = 72

Only variable in second degree is present.

Pure

vii. x^{2 }– x =0

Variable in both first and second degree is present

Adfected

**Question 2(i):**

Solve the quadratic equation:

x^{2 }– 196=0

**Solution :**

**Question 2(ii):**

Solve the quadratic equation:

5x^{2 }= 625

**Solution :**

**Question 2(iii):**

Solve the quadratic equation:

x^{2 }+ 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:

-4x^{2} + 324=0

**Solution :**

**Question 2(viii):**

Solve the quadratic equation:

-37.5x^{2}= -37.5

**Solution :**

**Question 3(i):**

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

x^{2 }+14x+13=0; x= – 1,x= – 13

**Solution :**

A number k satisfies the quadratic ax^{2}+bx+c=0 if

ak^{2}+bk+c=0

Given x^{2 }+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:

7x^{2 }– 12x = 0; x=1/3

**Solution :**

A number k satisfies the quadratic ax^{2}+bx+c=0 if

ak^{2}+bk+c = 0

Given 7x^{2 }– 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:

2m^{2 }– 6m + 3 = 0; m = 1/2

**Solution :**

A number k satisfies the quadratic ax^{2}+bx+c=0 if

ak^{2}+bk+c = 0

Given 2m^{2 }– 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 ax^{2}+bx+c=0 if

ak^{2}+bk+c=0

Given

Simplifying the equation we get

3x^{2} -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 ax^{2}+bx+c=0 if

ak^{2}+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:

6x^{2 }-x – 2 = 0; x = -1/2, x = 2/3

**Solution :**

A number k satisfies the quadratic ax^{2}+bx+c=0 if

ak^{2 }+ bk + c = 0

Given 6x^{2 }– 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 = πr^{2} solve for r, and find the value of r if A = 77 and π = 22/7.

**Solution :**

**Question 4(ii):**

If r^{2}= l^{2 }+ d^{2}, solve for d, and find the value of d if r = 5 and l = 4.

**Solution :**

**Question 4(iii):**

If c^{2}= a^{2}+b^{2}, 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 v^{2 }= u^{2 }+ 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:

x^{2 }+ 15x + 50 = 0

**Solution :**

x^{2 }+ 15x + 50 = 0

⟹ x^{2 }+ 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:

x^{2 }– 3x – 10 = 0

**Solution :**

x^{2 }– 3x – 10 = 0

⟹ x^{2 }+ 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 – p^{2 }= p

**Solution :**

6 – p^{2 }= p

⟹ p^{2 }+ p – 6 = 0

⟹ p^{2 }+ 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:

2x^{2 }+ 5x – 12 = 0

**Solution :**

2x^{2 }+ 5x – 12 = 0

⟹2x^{2 }+ 5x – 12 = 0

⟹2x^{2 }+ 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(m^{2 }+ 1)

**Solution :**

13m = 6(m^{2 }+ 1)

⟹6m^{2 }– 13m + 6 = 0

⟹6m^{2 }– 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:

100x^{2 }– 20x + 1 = 0

**Solution :**

100x^{2 }– 20x + 1= 0

⟹100x^{2 }– 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:

x^{2 }+ 4kx + 4k^{2 }= 0

**Solution :**

x^{2 }+ 4kx + 4k^{2 }= 0

⟹x^{2 }+ 2kx + 2kx + 4k^{2 }= 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 :**

⟹m^{2} – 6m – 7 = 0

⟹m^{2} – 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 :**

⟹2x^{2 }– 5x + 2 = 0

⟹2x^{2 }– 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:

21y^{2 }= 62y + 3

**Solution :**

21y^{2 }= 62y + 3

⟹21y^{2 }– 62y – 3 = 0

⟹21y^{2 }– 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:

0.2t^{2 }– 0.04t = 0.03

**Solution :**

0.2t^{2 }– 0.04t = 0.03

⟹0.2t^{2 }– 0.04t – 0.03 = 0

⟹Multiplying by 100 on both sides

⟹20t^{2 }– 4t – 3 = 0

⟹20t^{2 }– 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:

4x^{2 }+ 32x + 64 = 0

**Solution :**

4x^{2 }+ 32x + 64 = 0

Divide by 4 throughout the equation

⟹x^{2}+ 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:

a^{2}b^{2}x^{2 }– (a^{2 }+ b^{2})x + 1 = 0

**Solution :**

a^{2}b^{2}x^{2 }– (a^{2 }+ b^{2})x + 1 = 0

⟹a^{2}b^{2}x^{2 }– a^{2}x – b^{2}x + 1= 0

⟹a^{2}x(b^{2}x – 1) – 1(b^{2}x – 1) = 0

⟹(b^{2}x – 1)(a^{2}x – 1) = 0

⟹(b^{2}x – 1) = 0 or (a^{2}x – 1) = 0

⟹x = 1/b^{2} or x = 1/a^{2}

**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(x^{2 }+ 2x + 1) – 5x – 5 – 12 = 0

⟹2x^{2 }– x – 15 = 0

⟹2x^{2 }– 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}+12^{2 }= 15^{2}

**Solution :**

(x – 4)^{2 }+ 12^{2 }= 15^{2}⟹x^{2 }– 8x + 16 + 144 = 225

⟹x^{2 }– 8x + 65 = 0

⟹x^{2 }– 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:

4x^{2 }– 20x + 9 = 0

**Solution :**

4x^{2 }– 20x + 9 = 0

Dividing by coefficient of x^{2} on both sides

⟹ x^{2 }– 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:

4x^{2 }+ x – 5 = 0

**Solution :**

4x^{2 }+ x – 5 = 0

Dividing by coefficient of x^{2} on both sides

⟹ x^{2 }+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:

2x^{2 }+ 5x – 3 = 0

**Solution :**

2x^{2 }+ 5x – 3 = 0

Dividing by coefficient of x^{2} on both sides

⟹ x^{2 }+ 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:

x^{2 }+ 16x – 9 = 0

**Solution :**

x^{2 }+ 16x – 9 = 0

Adding and subtracting the square of half the coefficient of x

**Question 5:**

Solve the quadratic equation by completing the square:

x^{2 }– 3x + 1= 0

**Solution :**

x^{2 }– 3x + 1 = 0

Adding and subtracting the square of half the coefficient of x

**Question 6:**

Solve the quadratic equation by completing the square:

t^{2 }+ 3t = 7

**Solution :**

t^{2 }+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)

⟹ 3x^{2 }– 15x = 2x^{2 }+ 14x

⟹ x^{2 }– 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:

a^{2}x^{2 }– 3abx + 2b^{2 }= 0

**Solution :**

a^{2}x^{2 }– 3abx + 2b^{2}Dividing by coefficient of x^{2} on both sides

**Question 10:**

Solve the quadratic equation by completing the square:

4x^{2 }+ 4bx – (a^{2 }– b^{2}) = 0

**Solution :**

4x^{2 }+ 4bx – (a^{2 }– b^{2}) = 0

Dividing by coefficient of x^{2} on both sides

⟹ x^{2 }+ bx – (a^{2 }– b^{2})/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:

x^{2 }– 4x + 2=0

**Solution :**

**Question 2:**

Solve the quadratic equation by formula method:

x^{2 }– 2x + 4 = 0

**Solution :**

**Question 3:**

Solve the quadratic equation by formula method:

x^{2 }– 7x + 12 = 0

**Solution :**

**Question 4:**

Solve the quadratic equation by formula method:

2y^{2 }+ 6y = 3

**Solution :**

**Question 5:**

Solve the quadratic equation by formula method:

5m^{2 }– 11m + 2 = 0

**Solution :**

**Question 6:**

Solve the quadratic equation by formula method:

8r^{2 }= r + 2

**Solution :**

**Question 7:**

Solve the quadratic equation by formula method:

p = 5 – 2p^{2}

**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:

y^{2 }– 7y + 2 = 0

**Solution :**

**Question A(ii):**

Discuss the nature of the roots of the equation:

x^{2 }-2x+3=0

**Solution :**

**Question A(iii):**

Discuss the nature of the roots of the equation:

2n^{2 }+5n-1=0c

**Solution :**

**Question A(iv):**

Discuss the nature of the roots of the equation:

a^{2 }+4a+4=0

**Solution :**

**Question A(v):**

Discuss the nature of the roots of the equation:

x^{2 }+3x-4=0

**Solution :**

**Question A(vi):**

Discuss the nature of the roots of the equation:

3d^{2 }– 2d + 1= 0

**Solution :**

**Question B(i):**

For what positive values of m are the roots of the equation a^{2 }– ma + 1,

1.Equal 2.Distinct 3.Imaginary

**Solution :**

**Question B(ii):**

For what positive values of m are the roots of the equation x^{2}-mx+9,

1.Equal 2.Distinct 3.Imaginary

**Solution :**

**Question B(iii):**

For what positive values of m are the roots of the equation r^{2 }– (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 mk^{2}-3k+1,

1.Equal 2.Distinct 3.Imaginary

**Solution :**

**Question C(i):**

Find the values of p for which roots of the equation

x^{2} – px + 9 = 0 are equal.

**Solution :**

**Question C(ii):**

Find the values of p for which roots of the equation

2a^{2}+3a+p =0 are equal

**Solution :**

**Question C(iii):**

Find the values of p for which roots of the equation

pk^{2}-12k+9=0 are equal

**Solution :**

**Question C(iv):**

Find the values of p for which roots of the equation

2y^{2}-py+1=0 are equal

**Solution :**

**Question C(v):**

Find the values of p for which roots of the equation

(p+1)n^{2}+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)c^{2}+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:

x^{2 }-5x+8=0

**Solution :**

This is in the form of ax^{2}+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:

3a^{2 }– 10a – 5 = 0

**Solution :**

This is in the form of pa^{2 }+ 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:

8m^{2 }-m=2

**Solution :**

8m^{2 }-m=2

8m^{2 }– m – 2 = 0

This is in the form of am^{2 }+ 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:

6k^{2 }– 3 = 0

**Solution :**

This is in the form of ak^{2}+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:

pr^{2 }– r – 5 = 0

**Solution :**

This is in the form of ar^{2}+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:

x^{2 }+ (ab)x + (a+b)=0

**Solution :**

This is in the form of px^{2 }+ 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 x^{2} – (m + n)x + mn = 0

∴ The required quadratic equation is x^{2} – 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 x^{2} – (m + n)x + mn = 0

∴ The required quadratic equation is x^{2} – 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 x^{2} – 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 3m^{2} = 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 2a^{2} – 4a + 1 = 0. Find the value of

i. (p + q)^{2}+ 4pq

ii. p^{3}+ q^{3}

**Solution :**

**Question B(4):**

**Solution :**

**Question B(5):**

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

**Solution :**

Given equation is x^{2} + 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 2x^{2} – 3qx + 5q = 0 has one root which is twice the other.

**Solution :**

Given equation is 2x^{2} – 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 4x^{2} – 8px + 9 = 0 has roots whose difference is 4.

**Solution :**

Given equation is 4x^{2} – 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 x^{2} + px + q = 0 is 3 times the other prove that 3p^{2} = 16q.

**Solution :**

Given equation is x^{2} + 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 = -x^{2}

**Solution :**

y = -x^{2}

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | -9 | -4 | -1 | 0 | -1 | -4 | -9 |

**Question I(ii):**

Draw the graph of the following quadratic equation:

y = 3x^{2}

**Solution :**

y = 3x^{2}

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | 27 | 12 | 3 | 0 | 3 | 12 | 27 |

**Question I(iii):**

Draw the graph of the following quadratic equation:

y = x^{2 }+ 6x

**Solution :**

y = x^{2 }+ 6x

x | -7 | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 |

y | 7 | 0 | -5 | -8 | -9 | -8 | -5 | 0 | 7 |

**Question I(iv):**

Draw the graph of the following quadratic equation:

y = x^{2} – 2x

**Solution :**

y = x^{2} – 2x

x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

y | 8 | 3 | 0 | -1 | 0 | 3 | 8 |

**Question I(v):**

Draw the graph of the following quadratic equation:

y= x^{2} – 8x + 7

**Solution :**

y= x^{2} – 8x + 7

x | -1 | 0 | 1 | 2 | 4 | 7 | 8 |

y | 16 | 7 | 0 | -5 | -9 | 0 | 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 – x^{2} + 4 – 2x

∴ y = -x^{2} + 4

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | -5 | 0 | 3 | 4 | 3 | 0 | -5 |

**Question I(vii):**

Draw the graph of the following quadratic equation:

y = x^{2} + x – 6

**Solution :**

y = x^{2} + x – 6

x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | 6 | 0 | -4 | -6 | -6 | -4 | 0 | 6 |

**Question I(viii):**

Draw the graph of the following quadratic equation:

y = x^{2} – 2x + 5

**Solution :**

y = x^{2} – 2x + 5

x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

y | 13 | 8 | 5 | 4 | 5 | 8 | 13 |

**Exercise 9.10:**

**Question I(i):**

Draw the graph of the following equation.

y = x^{2}

**Solution :**

y = x^{2}

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | 9 | 4 | 1 | 0 | 1 | 4 | 9 |

**Question I(ii):**

Draw the graph of the following equation.

y = 3x^{2}

**Solution :**

y = 3x^{2}

x | -2 | -1 | 0 | 1 | 2 |

y | 12 | 3 | 0 | 3 | 12 |

**Question I(iii):**

Draw the graph of the following equation.

y = x^{2 } – 4x

**Solution :**

y = x^{2 } – 4x

x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |

y | 5 | 0 | -3 | -4 | -3 | 0 | 5 |

**Question I(iv):**

Draw the graph of the following equation.

y = -x^{2} + 8x – 16

**Solution :**

y = -x^{2} + 8x – 16

x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

y | -9 | -4 | -1 | 0 | -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).

∴ x^{2} + (x + 2)^{2} = 130

∴ x^{2} + x^{2} + 4x + 4 = 130

∴ 2x^{2} + 4x + 4 – 130 = 0

∴ 2x^{2} + 4x – 126 = 0

Dividing by 2 throughout, we get

x^{2} + 2x – 63 = 0

∴ x^{2} + 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.

∴ 3x^{2} – 4x = 15

∴ 3x^{2} – 4x – 15 = 0

∴ 3x^{2} – 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 + y^{2} = 12

∴ y^{2} + 4y – 12 = 0

∴ y^{2} + 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).

∴ x^{2} + (x + 1)(x + 2) = 154

∴ x^{2} + x^{2} + 2x + x + 2 – 154 = 0

∴ 2x^{2} + 3x – 152 = 0

∴ 2x^{2} + 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 + x^{2} – 304 = 0

∴ x^{2} + 25x – 150 = 0

∴ x^{2} + 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 (2x^{2}) years.

After eight years,

Son’s age = (x + 8) years

Father’s age = (2x^{2} + 8) years

According to given information, we have

(2x^{2} + 8) = 3(x + 8) + 4

∴ 2x^{2} + 8 = 3x + 24 + 4

∴ 2x^{2} – 3x + 8 – 28 = 0

∴ 2x^{2} – 3x – 20 = 0

∴ 2x^{2} – 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

∴ 2x^{2} = 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 = x^{2} – (5)^{2}∴ 56 = x^{2} – 25

∴ x^{2} = 56 + 25

∴ x^{2} = 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 cm^{2}. 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 = x^{2} + 6x

∴ x^{2} + 6x – 216 = 0

∴ x^{2} + 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

AB^{2} = AE^{2} + BE^{2}∴ (x + 8)^{2} = x^{2} + (x + 7)^{2}∴ x^{2 }+ 16x + 64 = x^{2} + x^{2} + 14x + 49

∴ x^{2} + 14x + 49 – 16x – 64 = 0

∴ x^{2} – 2x – 15 = 0

∴ x^{2} – 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 cm^{2}. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm^{2}. 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 = x^{2} Area of the bigger square = y^{2}According to given information, we have

y^{2} – 2x^{2} = 14

∴ y^{2} = 14 + 2x^{2} ….(1)

Also, 2y^{2} + 3x^{2} = 203 ….(2)

Substituting (1) in (2), we have

2(14 + 2x^{2}) + 3x^{2} = 203

∴ 28 + 4x^{2} + 3x^{2} – 203 = 0

∴ 7x^{2} – 175 = 0

∴ 7x^{2} = 175

Since side of a square cannot be negative, x = 5

∴ y^{2} = 14 + 2x^{2 }= 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,

BC^{2} = BD^{2} + DC^{2}(2x + 1)^{2} = (2x – 1)^{2} + x^{2}∴ 4x^{2} + 4x + 1 = 4x^{2} – 4x + 1 + x^{2}∴ x^{2} + 4x^{2} – 4x^{2} – 4x – 4x + 1 – 1 = 0

∴ x^{2} – 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.

∴ x^{2} = 100(24 – x)

∴ x^{2} = 2400 – 100x

∴ x^{2} + 100x – 2400 = 0

∴ x^{2} + 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(x^{2} – 6x)

∴ 8x – 24 = x^{2} – 6x

∴ x^{2} – 6x – 8x + 24 = 0

∴ x^{2} – 14x + 24 = 0

∴ x^{2} – 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 – 5t^{2}) 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 – 5t^{2} = 15

∴ 5t^{2} – 20t + 15 = 0

Dividing throughout by 5, we get

t^{2} – 4t + 3 = 0

∴ t^{2} – 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**

—————————————————————————–**All Subject KSEEB Solutions For Class 10**

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