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 โ 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} โ 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} โ 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 = โ 8) are real numbers.

**Question 1(iii):**

Check whether the following is quadratic equation:

x^{2 }+ ยฝ 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 = ยฝ, c = 0) are real numbers.

**Question 1(iv):**

Check whether the following is quadratic equation:

3x โ 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 = โ 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 โ 4)(2x โ 3) = 0

**Solution :**

(x โ 4)(2x โ 3) = 0

2x^{2 } โ 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 = โ 11 , c = 12 are real numbers.

**Question 2(iii):**

Simplify and check whether the following is quadratic equation:

(x + 9)(x โ 9) = 0

**Solution :**

(x + 9)(x โ 9) = 0

x^{2 }โ 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= โ 81) are real numbers.

**Question 2(iv):**

Simplify and check whether the following is quadratic equation:

(x + 2)(x โ 7) = 5

**Solution :**

( x + 2)(x โ 7) = 5

โนx^{2} โ 5x โ 14 โ 5= 0

โนx^{2} โ 5x โ 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 โ 1)(x โ 9) = 0

**Solution :**

3x + 2x^{2 }โ 19x + 9 = 0

2x^{2 }โ 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 โ 3)

**Solution :**

x^{2 }+ 2x + 1 = 2x โ 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 โ 1)(x โ 3) = (x + 5)(x โ 1)

**Solution :**

2x^{2 }โ 7x + 3 = x^{2 }+ 4x โ 5

x^{2 }โ 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 = โ 11 , c = 8) are real numbers.

**Question 2(viii):**

Simplify and check whether the following is quadratic equation:

x^{2 }+ 3x + 1 = (x โ 2)^{2}

**Solution :**

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

7x โ 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 }โ 1)

**Solution :**

(x + 2)^{3} = 2x(x^{2 }โ 1)

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

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

x^{3 }โ 6x^{2 }โ 14x โ 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} โ 4x^{2} โ x + 1 = (x โ 2)^{3}

**Solution :**

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

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

10x^{2 }โ 11x โ 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= โ 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 โ 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 โ 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 ร time

Given : speed = (x โ 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 โ 3) = 1

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

v. (x + 9)(x โ 9)

vi. 2x^{2} = 72

vii. x^{2 }โ 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 โ 3)=1

p^{2 }โ 3p โ 1=0

Variable in both first and second degree is present

Adfected

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

x^{2 }โ 2x + 3 =0

Variable in both first and second degree is present

Adfected

v. (x + 9)(x โ 9)

x^{2 }โ 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 }โ x =0

Variable in both first and second degree is present

Adfected

**Question 2(i):**

Solve the quadratic equation:

x^{2 }โ 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} โ 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= โ 1,x= โ 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 }โ 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 }โ 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 }โ 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 }โ 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 }โ 2(1) โ 1 = 3 โ 2 โ 1 = 0

Therefore, x = 1 satisfies the equation

Checking for x = โ 1

3( โ 1)^{2 }โ 2(- 1) โ 1 = 3 + 2 โ 1 = 4

Therefore, x = โ 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 โ 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 }โ x โ 2 = 0; x = -1/2, x = 2/3

Checking for x = โ 1/2

6(-1/2)^{2} โ (-1/2) โ 2 = 3/2 + 1/2 โ 2 = 0

Therefore, x = -1/2 satisfies the equation

Checking for x = 2/3

6(2/3)^{2} โ (2/3) โ 2 = 8/3 โ 2/3 โ 2 = 0

Therefore, x = 2/3 satisfies the equation.

**Question 4(i):**

If A = ฯr^{2} solve for r, and find the value of r if A = 77 and ฯ = 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

โน x^{2 }+ 5x + 10x + 50 = 0

โน x(x + 5) + 10(x + 5) = 0

โน (x + 10)(x + 5) = 0

โน x + 10 = 0 or x+5 = 0

โน x = โ 10 or x = -5

**Question 2:**

Solve the quadratic equation by factorization method:

x^{2 }โ 3x โ 10 = 0

**Solution :**

x^{2 }โ 3x โ 10 = 0

โน x^{2 }+ 2x โ 5x โ 10 = 0

โน (x + 2) โ 5(x + 2) = 0

โน (x + 2)(x โ 5) = 0

โน x + 2 = 0 or x โ 5 = 0

โน x = -2 or x = 5

**Question 3:**

Solve the quadratic equation by factorization method:

6 โ p^{2 }= p

**Solution :**

6 โ p^{2 }= p

โน p^{2 }+ p โ 6 = 0

โน p^{2 }+ 3p โ 2p โ 6 = 0

โนp(p + 3) โ 2(p + 3) = 0

โน(p โ 2)(p + 3) = 0

โนp โ 2 = 0 or p + 3 = 0

โนp = 2 or p = -3

**Question 4:**

Solve the quadratic equation by factorization method:

2x^{2 }+ 5x โ 12 = 0

**Solution :**

2x^{2 }+ 5x โ 12 = 0

โน2x^{2 }+ 5x โ 12 = 0

โน2x^{2 }+ 8x โ 3x โ 12 = 0

โน2x(x + 4) โ 3(x + 4) = 0

โน(2x โ 3)(x + 4) = 0

โน(2x โ 3) = 0 or (x + 4) = 0

โนx = 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)

โน6m^{2 }โ 13m + 6 = 0

โน6m^{2 }โ 9m โ 4m + 6 = 0

โน3m(2m โ 3) โ 2(2m โ 3) = 0

โน(3m โ 2)(2m โ 3) = 0

โน3m โ 2 = 0 or 2m โ 3 = 0

โนm = 2/3 or m = 3/2

**Question 6:**

Solve the quadratic equation by factorization method:

100x^{2 }โ 20x + 1 = 0

**Solution :**

100x^{2 }โ 20x + 1= 0

โน100x^{2 }โ 10x โ 10x + 1= 0

โน10x(10x โ 1) โ 1(10x โ 1) = 0

โน(10x โ 1)(10x โ 1) = 0

โน(10x โ 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

โนx^{2 }+ 2kx + 2kx + 4k^{2 }= 0

โนx(x + 2k) + 2k(x + 2k) = 0

โน(x + 2k)(x + 2k) = 0

โน(x + 2k)^{2 }=0

x = -2k is a repeated root of the equation

**Question 9:**

Solve the quadratic equation by factorization method:

**Solution :**

โนm^{2} โ 6m โ 7 = 0

โนm^{2} โ 7m + m โ 7 = 0

โนm(m โ 7) + 1(m โ 7) = 0

โน(m + 1)(m โ 7) = 0

โนm + 1= 0 or m-7=0

โนm = -1 or m = 7

**Question 10:**

Solve the quadratic equation by factorization method:

**Solution :**

โน2x^{2 }โ 5x + 2 = 0

โน2x^{2 }โ 4x โ x + 2 = 0

โน2x(x โ 2) โ 1(x โ 2) = 0

โน(2x โ 1)(x โ 2) = 0

โน2x โ 1 = 0 or x โ 2 = 0

โนx = 1/2 or x = 2

**Question 11:**

Solve the quadratic equation by factorization method:

21y^{2 }= 62y + 3

**Solution :**

21y^{2 }= 62y + 3

โน21y^{2 }โ 62y โ 3 = 0

โน21y^{2 }โ 63y + y-3 = 0

โน21y(y โ 3) + 1(y โ 3) = 0

โน(21y + 1)(y โ 3) = 0

โน21y + 1= 0 or y โ 3 = 0

โนy = -1/21 or y = 3

**Question 12:**

Solve the quadratic equation by factorization method:

0.2t^{2 }โ 0.04t = 0.03

**Solution :**

0.2t^{2 }โ 0.04t = 0.03

โน0.2t^{2 }โ 0.04t โ 0.03 = 0

โนMultiplying by 100 on both sides

โน20t^{2 }โ 4t โ 3 = 0

โน20t^{2 }โ 10t + 6t โ 3 = 0

โน10t(2t โ 1) + 3(2t โ 1)=0

โน10t + 3 = 0 or 2t โ 1= 0

โนt = -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

โนx^{2}+ 8x +16 = 0

โน(x + 4)^{2}= 0

โนx + 4 = 0

โนx = โ 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 }โ (a^{2 }+ b^{2})x + 1 = 0

**Solution :**

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

โนa^{2}b^{2}x^{2 }โ a^{2}x โ b^{2}x + 1= 0

โนa^{2}x(b^{2}x โ 1) โ 1(b^{2}x โ 1) = 0

โน(b^{2}x โ 1)(a^{2}x โ 1) = 0

โน(b^{2}x โ 1) = 0 or (a^{2}x โ 1) = 0

โนx = 1/b^{2} or x = 1/a^{2}

**Question 18:**

Solve the quadratic equation by factorization method:

2(x + 1)^{2 }โ 5(x + 1) = 12

**Solution :**

2(x + 1)^{2 }โ 5(x + 1) = 12

โน2(x^{2 }+ 2x + 1) โ 5x โ 5 โ 12 = 0

โน2x^{2 }โ x โ 15 = 0

โน2x^{2 }โ 6x + 5x โ 15 = 0

โน2x(x โ 3) + 5(x โ 3) = 0

โน(2x + 5)(x โ 3) = 0

โนx = -5/2 or x = 3

**Question 19:.**

Solve the quadratic equation by factorization method:

(x โ 4)^{2}+12^{2 }= 15^{2}

**Solution :**

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

โนx^{2 }โ 8x + 65 = 0

โนx^{2 }โ 13x + 5x โ 65 = 0

โนx(x โ 13) + 5(x โ 13) = 0

โน(x + 5)(x โ 13) = 0

โนx = โ 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 }โ 20x + 9 = 0

**Solution :**

4x^{2 }โ 20x + 9 = 0

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

โน x^{2 }โ 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 โ 5 = 0

**Solution :**

4x^{2 }+ x โ 5 = 0

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

โน x^{2 }+1/4x โ 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 โ 3 = 0

**Solution :**

2x^{2 }+ 5x โ 3 = 0

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

โน x^{2 }+ 5/2 x โ 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 โ 9 = 0

**Solution :**

x^{2 }+ 16x โ 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 }โ 3x + 1= 0

**Solution :**

x^{2 }โ 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 โ 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 โ 5) = 2x(x + 7)

**Solution :**

3x(x โ 5) = 2x(x + 7)

โน 3x^{2 }โ 15x = 2x^{2 }+ 14x

โน x^{2 }โ 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 }โ 3abx + 2b^{2 }= 0

**Solution :**

a^{2}x^{2 }โ 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 โ (a^{2 }โ b^{2}) = 0

**Solution :**

4x^{2 }+ 4bx โ (a^{2 }โ b^{2}) = 0

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

โน x^{2 }+ bx โ (a^{2 }โ 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 }โ 4x + 2=0

**Solution :**

**Question 2:**

Solve the quadratic equation by formula method:

x^{2 }โ 2x + 4 = 0

**Solution :**

**Question 3:**

Solve the quadratic equation by formula method:

x^{2 }โ 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 }โ 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 โ 2p^{2}

**Solution :**

**Question 8:**

Solve the quadratic equation by formula method:

(2x + 3)(3x โ 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 }โ 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 }โ 2d + 1= 0

**Solution :**

**Question B(i):**

For what positive values of m are the roots of the equation a^{2 }โ 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 }โ (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} โ 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 }โ 10a โ 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 }โ m โ 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 }โ 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 }โ r โ 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.

โด m = 3 and n = 5

Sum of the roots = m + n = 3 + 5 = 8

Product of the roots = mn = 3 ร 5 = 15

Standard form is x^{2} โ (m + n)x + mn = 0

โด The required quadratic equation is x^{2} โ 8x + 15 = 0.

**Question A(ii):**

Form the equation whose roots are 6, -5.

**Solution :**

Let m and n be the roots.

โด m = 6 and n = -5

Sum of the roots = m + n = 6 + (-5) = 1

Product of the roots = mn = 6 ร (-5) = -30

Standard form is x^{2} โ (m + n)x + mn = 0

โด The required quadratic equation is x^{2} โ x โ 30 = 0.

**Question A(iii):**

**Solution :**

**Question A(iv):**

**Solution :**

**Question A(v):**

**Solution :**

**Question A(vi):**

**Solution :**

**Question B(1):**

If โmโ and โnโ are the roots of the equation x^{2} โ 6x + 2 = 0, find the value of

i. (m + n)mn

**Solution :**

**Question B(2):**

If โaโ and โbโ are the roots of the equation 3m^{2} = 6m + 5, find the value of

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

**Solution :**

**Question B(3):**

If โpโ and โqโ are the roots of the equation 2a^{2} โ 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 โkโ 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.

โด a = 1, b = 4, c = (k + 2)

Let one root of the equation be โmโ and other root is zero.

**Question B(6):**

Find the value of โqโ so that the equation 2x^{2} โ 3qx + 5q = 0 has one root which is twice the other.

**Solution :**

Given equation is 2x^{2} โ 3qx + 5q = 0

โด a = 2, b = -3q and c = 5q

Let one root of the equation be โmโ.

Then the other root will be 2m.

**Question B(7):**

Find the value of โpโ so that the equation 4x^{2} โ 8px + 9 = 0 has roots whose difference is 4.

**Solution :**

Given equation is 4x^{2} โ 8px + 9 = 0

โด a = 4, b = -8p, c = 9

Let one root of the equation be โmโ.

Then, the other root will be (m โ 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

โด a = 1, b = p, c = q

Let one root of the equation be โmโ.

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} โ 2x

**Solution :**

y = x^{2} โ 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} โ 8x + 7

**Solution :**

y= x^{2} โ 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 โ x)

**Solution :**

y = (x + 2)(2 โ x)

โด y = 2x โ x^{2} + 4 โ 2x

โด 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 โ 6

**Solution :**

y = x^{2} + x โ 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} โ 2x + 5

**Solution :**

y = x^{2} โ 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 } โ 4x

**Solution :**

y = x^{2 } โ 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 โ 16

**Solution :**

y = -x^{2} + 8x โ 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).

โด x^{2} + (x + 2)^{2} = 130

โด x^{2} + x^{2} + 4x + 4 = 130

โด 2x^{2} + 4x + 4 โ 130 = 0

โด 2x^{2} + 4x โ 126 = 0

Dividing by 2 throughout, we get

x^{2} + 2x โ 63 = 0

โด x^{2} + 9x โ 7x โ 63 = 0

โด x(x + 9) โ 7(x + 9) = 0

โด (x + 9)(x โ 7) = 0

โด x = -9 or x = 7

But x is positive odd number.

โด 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.

โด 3x^{2} โ 4x = 15

โด 3x^{2} โ 4x โ 15 = 0

โด 3x^{2} โ 9x + 5x โ 15 = 0

โด 3x(x โ 3) + 5(x โ 3) = 0

โด (x โ 3)(3x + 5) = 0

โด 3x + 5 = 0 or x โ 3 = 0

โด 3x = -5 or x = 3

But x is a whole number.

โด 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.

โด x + y = 8

โด x = 8 โ y โฆ.(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 ร y = 12 โฆ.(1)

Again, the number formed by interchanging the digits is 10x + y.

According to given information, we have

10y + x + 36 = 10x + y

โด 10x + y โ x โ 10y = 36

โด 9x โ 9y = 36

Dividing by 9 throughout, we get

x โ y = 4

โด x = 4 + y โฆ.(2)

Substituting (2) in (1), we get

(4 + y) ร y = 12

โด 4y + y^{2} = 12

โด y^{2} + 4y โ 12 = 0

โด y^{2} + 6y โ 2y โ 12 = 0

โด y(y + 6) โ 2(y + 6) = 0

โด (y + 6)(y โ 2) = 0

โด y + 6 = 0 or y โ 2 = 0

โด y = -6 or y = 2

Since digit cannot be negative, y = 2

โด 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).

โด x^{2} + (x + 1)(x + 2) = 154

โด x^{2} + x^{2} + 2x + x + 2 โ 154 = 0

โด 2x^{2} + 3x โ 152 = 0

โด 2x^{2} + 19x โ 16x โ 152 = 0

โด x(2x + 19) โ 8(2x + 19) = 0

โด (2x + 19)(x โ 8) = 0

โด 2x + 19 = 0 or x โ 8 = 0

โด 2x = -19 or x = 8

Since x is a positive integer, x = 8

โด 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,

Kavyaโs age = (11 + x) years

Karthikโs age = (14 + x) years

โด (11 + x)(14 + x) = 304

โด 154 + 11x + 14x + x^{2} โ 304 = 0

โด x^{2} + 25x โ 150 = 0

โด x^{2} + 30x โ 5x โ 150 = 0

โด x(x + 30) โ 5(x + 30) = 0

โด (x + 30)(x โ 5) = 0

โด x + 30 = 0 or x โ 5 = 0

โด 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,

Sonโs age = (x + 8) years

Fatherโs age = (2x^{2} + 8) years

According to given information, we have

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

โด 2x^{2} + 8 = 3x + 24 + 4

โด 2x^{2} โ 3x + 8 โ 28 = 0

โด 2x^{2} โ 3x โ 20 = 0

โด 2x^{2} โ 8x + 5x โ 20 = 0

โด 2x(x โ 4) + 5(x โ 4) = 0

โด (x โ 4)(2x + 5) = 0

โด x โ 4 = 0 or 2x + 5 = 0

โด x = 4 or 2x = -5

Since, age cannot be negative, x = 4

โด 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 โ 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 โ 5

Area of a rectangle = 56 cm

Now, area of a rectangle = Length ร Breadth

โด 56 = (x + 5)(x โ 5)

โด 56 = x^{2} โ (5)^{2}โด 56 = x^{2} โ 25

โด x^{2} = 56 + 25

โด x^{2} = 81

โด x = ยฑ9

Since the dimensions of a rectangle cannot be negative, x = 9.

โด Base of a rectangle = x + 5 = 9 + 5 = 14 cm

Height of a rectangle = x โ 5 = 9 โ 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 โxโ cm.

Then, altitude (height) of a triangle = (x + 6) cm

โด 216 = x^{2} + 6x

โด x^{2} + 6x โ 216 = 0

โด x^{2} + 18x โ 12x โ 216 = 0

โด x(x + 18) โ 12(x + 18) = 0

โด (x + 18)(x โ 12) = 0

โด x = -18 or x = 12

Since the base of a triangle cannot be negative, x = 12.

โด 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 โAEB, by using Pythagoras theorem, we have

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

โด x^{2} + 14x + 49 โ 16x โ 64 = 0

โด x^{2} โ 2x โ 15 = 0

โด x^{2} โ 5x + 3x โ 15 = 0

โด x(x โ 5) + 3(x โ 5) = 0

โด (x โ 5)(x + 3) = 0

โด x โ 5 = 0 or x + 3 = 0

โด x = 5 or x = -3

Since length cannot be negative, x = 5.

โด Diagonal AC = AE + EC = x + x = 2x = 2 ร 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 โxโ cm and side of the bigger square be โyโ cm.

โด Area of the smaller square = x^{2} Area of the bigger square = y^{2}According to given information, we have

y^{2} โ 2x^{2} = 14

โด y^{2} = 14 + 2x^{2} โฆ.(1)

Also, 2y^{2} + 3x^{2} = 203 โฆ.(2)

Substituting (1) in (2), we have

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

โด 28 + 4x^{2} + 3x^{2} โ 203 = 0

โด 7x^{2} โ 175 = 0

โด 7x^{2} = 175

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

โด y^{2} = 14 + 2x^{2 }= 14 + 2(5)^{2} = 14 + 50 = 64

โด y = ยฑ8

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 โ 1 and BC = 2x + 1, find the lengths of all three sides of the triangle.

**Solution :.**

Consider the following figure:

In โABC, BD โฅ AC

โด AD = DC = x

In right-angled โBDC, by Pythagoras theorem,

BC^{2} = BD^{2} + DC^{2}(2x + 1)^{2} = (2x โ 1)^{2} + x^{2}โด 4x^{2} + 4x + 1 = 4x^{2} โ 4x + 1 + x^{2}โด x^{2} + 4x^{2} โ 4x^{2} โ 4x โ 4x + 1 โ 1 = 0

โด x^{2} โ 8x = 0

โด x(x โ 8) = 0

โด x = 0 or x = 8

Since measurement cannot be negative, x = 8.

AC = AD + DC = x + x = 2x = 2 ร 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 โ 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.

โด x^{2} = 100(24 โ x)

โด x^{2} = 2400 โ 100x

โด x^{2} + 100x โ 2400 = 0

โด x^{2} + 120x โ 20x โ 2400 = 0

โด x(x + 120) โ 20(x + 120) = 0

โด (x + 120)(x โ 20) = 0

โด x + 120 = 0 or x โ 20 = 0

โด 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 โ 6) days.

>

โด 4(2x โ 6) = 1(x^{2} โ 6x)

โด 8x โ 24 = x^{2} โ 6x

โด x^{2} โ 6x โ 8x + 24 = 0

โด x^{2} โ 14x + 24 = 0

โด x^{2} โ 12x โ 2x + 24 = 0

โด x(x โ 12) โ 2(x โ 12) = 0

โด (x โ 12)(x โ 2) = 0

โด x โ 12 = 0 or x โ 2 = 0

โด x = 12 or x = 2

x = 2 has no meaning as 6 days less than 2 will be -4.

โด 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 โ 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 :**

โด t(20 โ 5t) = 15

โด 20t โ 5t^{2} = 15

โด 5t^{2} โ 20t + 15 = 0

Dividing throughout by 5, we get

t^{2} โ 4t + 3 = 0

โด t^{2} โ 3t โ t + 3 = 0

โด t(t โ 3) โ 1(t โ 3) = 0

โด (t โ 3)(t โ 1) = 0

โด t โ 3 = 0 or t โ 1 = 0

โด t = 3 or t = 1

When t = 3, speed = 20 โ 5t = 20 โ 5 ร 3 = 20 โ 15 = 5 m/s

When t = 1, speed = 20 โ 5t = 20 โ 5 ร 1 = 20 โ 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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