Important Questions Class 10 Maths Chapter 4

Important Questions Class 10 Mathematics Chapter 4 – Quadratic Equations

Mathematics requires one to use a lot of interpersonal and intrapersonal skills in order to perform well in their examinations. This can be built with regular practice and the right guidance from the experts in Mathematics and then taking crucial steps in the preparation.

The concepts you have studied in the chapter, Polynomials will be largely used in the chapter of Quadratic equations. Quadratic equations are the study of the various complex sets of equations that will be used in the higher classes. The important questions covered in the chapter include Quadratic equations, solutions of a quadratic equation by completing the square, and the nature of the roots with their examples.

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Important Questions Class 10 Mathematics Chapter 4 – With Solutions

The following important questions and their solutions are included in the Class 10 Mathematics Chapter. 

 

Question 1. Which of the following options is a quadratic equation?

(A) x2 + 2x + 1 = (4 – x)2 + 3

(B) –2x2 = (5 – x)(2x-(2/5))

(C) (k + 1) x2 + (3/2) x = 7, here k = –1 

(D) x3 – x2 = (x – 1)3

Answer 1:

(D) x3 – x2 = (x – 1)3

Explanation:

The standard form of a quadratic equation is given by,

ax2 + bx + c = 0, a ≠ 0

(A) Given that 

x2 + 2x + 1 = (4 – x)2 + 3

x2 + 2x + 1 = 16 – 8x + x2 + 3

10x – 18 = 0

it is not a quadratic equation.

(B) Given that,

-2x2 = (5 – x) (2x – 2/5)

-2x2 = 10x – 2x2 – 2 +2/5x

52x – 10 = 0

it is not a quadratic equation.

(C) Given that,

(k + 1) x2 + 3/2 x  = 7, here k = -1

(-1 + 1) x2 + 3/2 x = 7

3x – 14 = 0

it is not a quadratic equation.

(D) Given that,

x3 – x2 = (x – 1)3

x3 – x2 = x3 – 3x2 + 3x – 1

2x2 – 3x + 1 = 0

It is a quadratic equation.

 

Question 2. Which among the following options is not a quadratic equation?

(A) 2(x – 1)2 = 4x2 – 2x + 1 

(B) 2x – x2 = x2 + 5

(C) (√2x + √3)2 + x2 = 3x2 − 5x

(D) (x2 + 2x)2 = x4 + 3 + 4x3

Answer 2: (D) (x2 + 2x)2 = x4 + 3 + 4x3

Explanation :A quadratic equation is represented in the form,

ax2 + bx + c = 0, a ≠ 0

(A) Given that, 

2(x – 1)2 = 4x2 – 2x + 1

2(x2 – 2x + 1) = 4x2 – 2x + 1

2x2 + 2x – 1 = 0

It is a quadratic equation.

(B) Given that,

2x – x2 = x2 + 5

2x2 – 2x + 5 = 0

which is a quadratic equation.

(C) Given that, 

(√2x + √3)2  = 3x2 – 5x

2x2 + 2√6x + 3  = 3x2 – 5x

x2 – (5 + 2√6)x – 3 = 0

It is a quadratic equation.

(D) Given that, 

(x2 + 2x)2 = x4 + 3 + 4x2

x4 + 4x3 + 4x2 = x4 + 3 + 4x2

4x3 – 3 = 0

which is a cubic equation but not a quadratic equation.

 

Question 3. Which of the following equations contains 2 as a root?

(A) x2 – 4x + 5 = 0 

(B) x2 + 3x – 12 = 0

(C) 2x2 – 7x + 6 = 0 

(D) 3x2 – 6x – 2 = 0

Answer 3: (C) 2x2 – 7x + 6 = 0

Explanation : When 2 is a root, 

putting the value 2 in place of x should satisfy the equation.

(A) Given that,

x2 – 4x + 5 = 0

(2)2 – 4(2) + 5 = 1 ≠ 0

Thus, x = 2 is not a root of x2 – 4x + 5 = 0

(B) Given that, 

x2 + 3x – 12 = 0

(2)2 + 3(2) – 12 = -2 ≠ 0

Thus, 

x = 2 is not the root of x2 + 3x – 12 = 0

(C) as Given that 

2x2 – 7x + 6 = 0

2(2)2 – 7(2) + 6 = 0

Where, x = 2 is a root of 2x2 – 7x + 6 = 0

(D) Given that, 

3x2 – 6x – 2 = 0

3(2)2 – 6(2) – 2 = -2 ≠ 0

Thus, x = 2 is not a root of 3x2 – 6x – 2 = 0

 

Question 4. If ½ is a root of the equation x2 + kx – 5/4 = 0, the value of k is

(A) 2 

(B) – 2

(C) ¼ 

(D) ½

Answer 4: (A) 2

Explanation : If  ½ is a root of the equation x2 + kx – 5/4 = 0 

Putting the value of ½ in place of x should help us obtain the value of k.

Given that,

x2 + kx – 5/4 = 0 here, x = ½

(½)2 + k (½) – (5/4) = 0

(k/2) = (5/4) – ¼

k = 2

 

Question 5. Which of the following equations have the sum of its roots as 3?

(A) 2x2 – 3x + 6 = 0 

(B) –x2 + 3x – 3 = 0

(C) √2x2 – 3/√2x+1=0 

(D) 3x2 – 3x + 3 = 0

Answer 5:(B) –x2 + 3x – 3 = 0

Explanation : The sum of the roots for the quadratic equation ax2 + bx + c = 0, a ≠ 0 is given as,

Coefficient for x / coefficient for x2 = – (b/a)

(A) Given that, 

2x2 – 3x + 6 = 0

sum of the roots = – b/a = -(-3/2) = 3/2

(B) Given that, 

–x2 + 3x – 3 = 0

sum of the roots = – b/a = -(3/-1) = 3

(C) Given that, 

√2x2 – 3/√2x+1=0

2x2 – 3x + √2 = 0

sum of the roots = – b/a = -(-3/2) = 3/2

(D) Given that, 

3x2 – 3x + 3 = 0

sum of the roots = – b/a = -(-3/3) = 1

 

 

 

Question 6: Values of k for which the quadratic equation 2x2 – kx + k = 0 has equal roots is

(A) 0 only

(B) 4

(C) 8 only

(D) 0, 8

 

Answer 6: (D) 0, 8

Explanation :Given that,

2x2−kx+k=0

then, comparing with ax2+bx+c=0, we observe

a = 2, b = −k and c = k.

 

For equal roots, the discriminant might be zero.

so, we can say, D=b2−4ac=0.

(−k)2−4(2)k=0

k2−8k=0

k(k−8)=0.   

 

Hence, k = 0, 8

 

Hence, the required values of k are 0 and 8.

Therefore, the correct answer is option D.

 

 

 

 

Question 7: Which constant should be added and subtracted to solve the quadratic equation 9x2+3/4x−√2 =0 by the method of completing the square?

(A) 1/8

(B) 1/64

(C) 1/4

(D) 9/64

Answer 7: (B) 1/64

Explanation :Given, equation 9x2+3/4x−√2 =0.

⇒(3x)2+1/4(3x)−√2=0

Substituting 3x = y in the above equation, we observe

y2+1/4y–√2=0

y2+1/4y+(1/8)2−(1/8)2–√2=0⇒(y+1/8)2=1/64+√2

(y+1/8)2=(1+64√2)/64

 

Thus, 1/64 must be added and subtracted to solve the given equation.

Hence, the correct answer is option B.

 

 

 

Question 8: The quadratic equation 2x2−√5x+1=0 has

(A) two distinct real roots

(B) two equal real roots

(C) no real roots

(D) more than 2 real roots

 

Answer 8: (C) no real roots

Explanation : Given that, equation 2x2−√5x+1=0.

then, comparing with ax2+bx+c=0, we get

a = 2, b = −√5 and c = 1.

⇒D=b2−4ac=(−√5)2−4*2*1=5−8, here D is discriminant.

D =−3, that is less than 0.

 

Hence, we conclude a given quadratic equation has no real roots since the discriminant is negative.

Hence, the correct answer is option C

 

 

Question 9: Which among the following equations has two distinct real roots?

(A) 2x2−3√2x+9/4=0

(B) x2 + x – 5 = 0

(C) x2+ 3x+2√2 =0

(D) 5x2 – 3x + 1 = 0

 

Answer 9: (B) x2 + x – 5 = 0

Explanation: when a quadratic equation is in the form of ax2+bx+c=0, a≠0 then

(i) when D=b2−4ac>0, then its roots are distinct and real.

(ii) when D=b2−4ac=0, then its roots are real and equal.

(iii) when D=b2−4ac<0, then its roots are not real or imaginary roots.

(a) Given, 2x2−3√2–x+9/4=0.

then, comparing with ax2+bx+c=0, we get

a = 2, b = −3√2 and c = 9/4.

⇒D=b2−4ac=(−3√2)2−4×2×9/4=18−18=0, where D is discriminant.

D=0.

Thus, the equation has real and equal roots.

(b) Given x2+x−5=0.

then, comparing with ax2+bx+c=0, we observe

a = 1, b = 1 as well as c = −5.

⇒D=b2−4ac=12−4×1×(−5)=1+20=21, here D is discriminant.

⇒D=21, that is greater than zero.

Hence, the equation has real and distinct roots.

(c) Given x2+3x+2√2=0.

then, comparing with ax2+bx+c=0, we get

a = 1, b = 3 and c = 2√2

⇒D=b2−4ac=(3)2−4×1×(2√2)= 9−8√2<0, here D is discriminant.

⇒D<0

Hence, the equation has no real roots.

(d) Given 5x2−3x+1=0.

then, comparing with ax2+bx+c=0, we get

a = 5, b = −3 and c = 1

⇒D=b2−4ac=(−3)2−4×5×1=9−20<0, where D is discriminant.

D<0

Thus, the equation has no real roots.

Hence, the correct answer is option B.

 

 

 

 

Question 10: Which among the following equations has no real roots?

(A) x2−4x+3√2=0

(B) x2+4x−3√2=0

(C) x2−4x−3√2=0

(D) 3x2+4√3x+4=0

Answer 10: (A) x2−4x+3√2=0

Explanation: when a quadratic equation is in the form of ax2+bx+c=0, a≠0, then

(i) when D=b2−4ac>0, then its roots are distinct and real.

(ii) when D=b2−4ac=0, then its roots are real and equal.

(iii) when D=b2−4ac<0, then its roots are not real or imaginary roots.

(a) Given x2−4x+3√2=0.

then, comparing with ax2+bx+c=0, we observe

a = 1, b = −4 as well as c = 3√2

⇒D=b2−4ac

=(−4)2−4*1*3√2

=16−12√2<0, where D is discriminant.

⇒D<0.

Thus, the equation has no real roots.

(b) Given x2+4x−3√2=0.

then, comparing with ax2+bx+c=0, we observe

a = 1, b = 4 as well as c = −3√2.

⇒D=b2−4ac

=(4)2−4×1×(−3√2)

=16+12√2>0, where D is discriminant.

⇒D>0.

Thus, the equation has real and distinct roots.

(c) Given x2−4x−3√2=0.

then, comparing with ax2+bx+c=0, we observe

a = 1, b = −4 as well as c = −3√2

⇒D=b2−4ac

=(−4)2−4*1*(−3√2)  

=16+12√2>0, where D is discriminant.

⇒D>0.

Thus, the equation has real and distinct roots.

(d) Given 3x2+4√3x+4=0.

then, comparing with ax2+bx+c=0, we observe

a = 3, b = 4√3 as well as c = 4.

⇒D=b2−4ac

=(4√3)2−4*3*4

=48−48=0, where D is discriminant.

⇒D=0.

Thus, the equation has real and equal roots.

Hence, the correct answer is option (A).

 

Question 11: (x2 + 1)2 – x2 = 0 has

(A) four real roots

(B) two real roots

(C) no real roots

(D) one real root.

 

Answer 11: (C) no real roots

Explanation: Given equation, (x2+1)2−x2=0.

x4+1+2x2−x2=0

x4+x2+1=0

Let x2=y

 (x2)2+x2+1=0

y2+y+1=0

then, comparing with ay2+by+c=0, we get.

a = 1, b = 1 and c = 1

Hence, we can say D=b2−4ac, here D is discriminant.

=(1)2−4(1)(1)=1−4=−3

Since, D<0.

Hence, we conclude the given equation has no real roots.

Therefore, the correct answer is option C.

 

Question 12: State if the following quadratic equations have two distinct real roots. Give explanations from the answer.

1. x2 – 3x + 4 = 0
2. 2x2 + x – 1 = 0
3. 2x2 – 6x + 9/2 = 0
4. 3x2 – 4x + 1 = 0
5. (x + 4)2 – 8x = 0
6. (x – √2)2 – 2(x + 1) = 0
7. √2 x2 –(3/√2)x + 1/√2 = 0
8. x (1 – x) – 2 = 0
9. (x – 1) (x + 2) + 2 = 0
10. (x + 1) (x – 2) + x = 0

Answer 12:

(i) equation x2 – 3x + 4 = 0 gives no real roots,As,

D = b2 – 4ac

= (-3)2 – 4(1)(4)

= 9 – 16 < 0

Thus, the roots are imaginary.

(ii) equation 2x2 + x – 1 = 0 gives two real and distinct roots,As,

D = b2 – 4ac

= 12 – 4(2) (-1)

= 1 + 8 > 0

Thus, the roots are real and distinct.

(iii) equation 2x2 – 6x + (9/2) = 0 gives real and equal roots,As,

D = b2 – 4ac

= (-6)2 – 4(2) (9/2)

= 36 – 36 = 0

Thus, the roots are real and equal.

(iv) equation 3x2 – 4x + 1 = 0 gives two real and distinct roots,As

D = b2 – 4ac

= (-4)2 – 4(3)(1)

= 16 – 12 > 0

Thus, the roots are real and distinct.

(v) equation (x + 4)2 – 8x = 0 gives no real roots,As,

After simplifying the above equation, we get,

x2 + 8x + 16 – 8x = 0

x2 + 16 = 0

D = b2 – 4ac

= (0)2– 4(1) (16) < 0

Thus, the roots are imaginary.

(vi) equation (x – √2)2 – √2(x+1)=0 gives two distinct and real roots,As,

After simplifying the above equation, we get,

x2 – 2√2x + 2 – √2x – √2 = 0

x2 – √2(2+1)x + (2 – √2) = 0

x2 – 3√2x + (2 – √2) = 0

D = b2 – 4ac

= (– 3√2)2 – 4(1)(2 – √2)

= 18 – 8 + 4√2 > 0

Thus,the roots are real and distinct.

(vii) equation √2x2 – 3√2x + ½ = 0 contains two real and distinct roots.

D = b2 – 4ac

= (- 3/√2)2 – 4(√2) (½)

= (9/2) – 2√2 > 0

Thus, the roots are real and distinct.

(viii) equation x (1 – x) – 2 = 0 contains no real roots.

On simplifying the above equation, we get

x2 – x + 2 = 0

And

D = b2 – 4ac

= (-1)2 – 4(1)(2)

= 1 – 8 < 0

Thus, the roots are imaginary.

(ix) equation (x – 1) (x + 2) + 2 = 0 contains two real and distinct roots.

On simplifying the above equation, we get,

x2 – x + 2x – 2 + 2 = 0

x2 + x = 0

And

D = b2 – 4ac

= 12 – 4(1)(0)

= 1 – 0 > 0

Thus, the roots are real and distinct.

(x) equation (x + 1) (x – 2) + x = 0 contains two real and distinct roots.

On simplifying the above equation, we get,

x2 + x – 2x – 2 + x = 0

x2 – 2 = 0

And

D = b2 – 4ac

= (0)2 – 4(1) (-2)

= 0 + 8 > 0

Thus, the roots are real and distinct.

 

 

 

Question 13. Say whether the following statements are true or false. Explain your answers.

1. Every quadratic equation contains exactly one root.
2. Every quadratic equation contains at least one real root.
3. Every quadratic equation contains at least two roots.
4. Every quadratic equation contains at most two roots.
5. If the coefficient for x2 and the constant term for the quadratic equation have opposite signs, then the quadratic equation contains real roots.
6. If the coefficient for x2 and the constant term has the same sign, as well as when the coefficient of x term is zero, then the quadratic equation has no real roots.

Answer 13:

(i) False. 

For instance, a quadratic equation x2 – 9 = 0 contains two distinct roots – 3 and 3.

(ii) False. 

For instance, equation x2 + 4 = 0 contains no real root.

(iii) False. 

For instance, a quadratic equation x2 – 4x + 4 = 0 contains only one root which is 2.

(iv) True.

As every quadratic polynomial contains almost two roots.

(v) True.

As in such cases, the discriminant is always positive.

For instance, in ax2+ bx + c = 0, for a and c have opposite sign, ac < 0

Thus, Discriminant = b2 – 4ac > 0.

(vi) True.

As in such cases, discriminant is always negative.

For instance, ax2+ bx + c = 0, as b = 0, for a and c have same sign then ac > 0

Thus, Discriminant = b2 – 4ac = – 4 a c < 0

 

Question 14. The quadratic equation having an integral coefficient has integral roots. Give explanations for your answer.

Answer 14:

No, the quadratic equation with integral coefficients may or may not contain integral roots.

Explanation

For the following equation,

8x2 – 2x – 1 = 0

The roots of the equation are ½ and – ¼ that are not integers.

Thus, the quadratic equation with integral coefficient might or might not contain integral roots.

 

 

 

Question 15. Find out the roots for the quadratic equations given below by using the quadratic formula in each of them:

1. 2x2 – 3x – 5 = 0
2. 5x2 + 13x + 8 = 0
3. –3x2 + 5x + 12 = 0
4. –x2 + 7x – 10 = 0
5. x2 + 2 √2x – 6 = 0
6. x2 – 3 √5x + 10 = 0
7. (½)x2– √11x + 1 = 0

Answer 15:

The quadratic formula for finding the roots for the quadratic equation

ax2 + bx + c = 0, a ≠ 0 is represented by,

 

x = [-(b) ± √{(b)2 – 4(a)*(c)}] /2(a)

 

(i) 2x2 – 3x – 5 = 0

x = [-(-3) ± √{(3)2 – 4(2)*(-5)}] /2(2)

x = (3 ± √49)/4

x = (3 ± 7)/4 = -1, 5/2

(ii) 5x2 + 13x + 8 = 0

x = [-(13) ± √{(13)2 – 4(5)*(8)}] /2(5)

x = (-13 ± √9)/10

x = (-13 ± 3)/10 = -1, -8/5

(iii) –3x2 + 5x + 12 = 0

x = [-(5) ± √{(5)2 – 4(-3)*(12)}] /2(-3)

x = (-5 ± √169)/-6

x = (5 ± 13)/6 = 3, -4/3

(iv) –x2 + 7x – 10 = 0

x = [-(7) ± √{(7)2 – 4(-1)*(-10)}] /2(-1)

x = (-7 ± √9)/-2

x = (7 ± 3)/2 = 5, 2

(v) x2 + 2 √2x – 6 = 0

x = [-(2√2) ± √{(2√2)2 – 4(1)*(-6)}] /2(1)

x = (-2√2 ± √32)/2

x = (-2√2 ± 4√2)/2 = √2, -3√2

(vi) x2 – 3 √5x + 10 = 0

x = [-(-3√5) ± √{(-3√5)2 – 4(1)*10}] /2(1)

x = (3√5 ± √5)/2 = 2√5, √5

(vii) (½)x2– √11x + 1 = 0

x = [-(-√11) ± √{(-√11)2 – 4(1/2)*1}] /2(1/2)

x = (√11 ± √9)/1

x = √11 ± 3 = 3 + √11, -3 + √11

 

 

 

 

Question 16. Check if the following are quadratic equations:

(i) (x + 1)2 = 2(x – 3)

(ii) x2 – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x2 + 3x + 1 = (x – 2)2

(vii) (x + 2)3 = 2x (x2 – 1)

(viii) x3– 4x2 – x + 1 = (x – 2)3

Answer 16:

(i) Given that,

(x + 1)2 = 2(x – 3)

Using the formula, 

(a+b)2 = a2+2ab+b2

We get,

 x2 + 2x + 1 = 2x – 6

x2 + 7 = 0

As the above equation is in the form of ax2 + bx + c = 0.

Hence, the given equation is a quadratic equation.

(ii) Given that, 

x2 – 2x = (–2) (3 – x)

 x2 – 2x = -6 + 2x

 x2 – 4x + 6 = 0

As the above equation is in the form of ax2 + bx + c = 0.

Hence, the given equation is a quadratic equation.

(iii) Given that, 

(x – 2)(x + 1) = (x – 1)(x + 3)

Multiplying, we get,

 x2 – x – 2 = x2 + 2x – 3

 3x – 1 = 0

As the above equation is not in the form of ax2 + bx + c = 0.

so, the given equation is not a quadratic equation.

(iv) Given that, (x – 3)(2x +1) = x(x + 5)

Multiplying, we get,

 2x2 – 5x – 3 = x2 + 5x

x2 – 10x – 3 = 0

As the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is a quadratic equation.

(v) Given that, 

(2x – 1)(x – 3) = (x + 5)(x – 1)

Multiplying, we get,

2x2 – 7x + 3 = x2 + 4x – 5

x2 – 11x + 8 = 0

As the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is a quadratic equation.

(vi) Given that, 

x2 + 3x + 1 = (x – 2)2

 Using the formula, we get, (a-b)2=a2-2ab+b2

 x2 + 3x + 1 = x2 + 4 – 4x

7x – 3 = 0

As the above equation is not in the form of ax2 + bx + c = 0.

Thus, the given equation is not a quadratic equation.

(vii) Given that, 

(x + 2)3 = 2x(x2 – 1)

Using the formula, we get, (a+b)3 = a3+b3+3ab(a+b)

 x3 + 8 + x2 + 12x = 2x3 – 2x

x3 + 14x – 6x2 – 8 = 0

As the above equation is not in the form of ax2 + bx + c = 0.

Thus, the given equation is not a quadratic equation.

(viii) Given that,

x3 – 4x2 – x + 1 = (x – 2)3

 Using the formula, we get, (a-b)3 = a3-b3-3ab(a-b)

 x3 – 4x2 – x + 1 = x3 – 8 – 6x2  + 12x

 2x2 – 13x + 9 = 0

As the above equation is in the form of ax2 + bx + c = 0.

Hence, the given equation is a quadratic equation.

 

 

 

Question 17. Find out the roots of quadratic equations by factorisation:

(i) √2 x2 + 7x + 5√2=0

(ii) 100x2 – 20x + 1 = 0

Answer 17:

(i)  √2 x2 + 7x + 5√2=0

By considering the L.H.S. first,

⇒ √2 x2 + 5x + 2x + 5√2

⇒ x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of the equation, √2 x2 + 7x + 5√2=0 are the values of x which have (√2x + 5)(x + √2) = 0

Thus, √2x + 5 = 0 or x + √2 = 0

⇒ x = -5/√2 or x = -√2

(ii) Given that, 

100x2 – 20x + 1=0

By considering the L.H.S. first,

100x2 – 10x – 10x + 1

10x(10x – 1) -1(10x – 1)

 (10x – 1)2

The roots for this equation, 100x2 – 20x + 1=0, are the values of x  which have (10x – 1)2= 0

Thus,

(10x – 1) = 0

And 

(10x – 1) = 0

 x =1/10 or x =1/10

 

 

 

 

Question 18 Find out the two consecutive positive integers, the sum of whose squares is 365.

Answer 18: Assume the two consecutive positive integers be x and x + 1.

Hence, 

According to the given statement,

x2 + (x + 1)2 = 365

x2 + x2 + 1 + 2x = 365

2x2 + 2x – 364 = 0

 x2 + x – 182 = 0

 x2 + 14x – 13x – 182 = 0

 x(x + 14) -13(x + 14) = 0

 (x + 14)(x – 13) = 0

So, if, x + 14 = 0 or x – 13 = 0,

Then,  x = – 14 or x = 13

As the integers are positive, x can be 13, only.

Thus, x + 1 = 13 + 1 = 14

Hence, the two consecutive positive integers will be 13 and 14.

 

 

Question 19. Solve for the quadratic equation 2x2 – 7x + 3 = 0 by using the quadratic formula.

Answer 19: 

We have,

2x2 – 7x + 3 = 0

if we compare the given equation with ax2 + bx + c = 0, we find,

a = 2, b = -7 and c = 3

Using quadratic formula, we have

x = [-b±√(b2 – 4ac)]/2a

 x = [7±√(49 – 24)]/4

 x = [7±√25]/4

 x = [7±5]/4

Thus,

 x = 7+5/4 or x = 7-5/4

 x = 12/4 or 2/4

Therefore, x = 3 or ½

 

 

 

 

Question 20: Find out the values for k for each of the following quadratic equations, such that they have two equal roots.

(i) 2x2 + kx + 3 = 0

(ii) kx (x – 2) + 6 = 0

Answer 20:

(i) 2x2 + kx + 3 = 0

On comparing the given equation with ax2 + bx + c = 0, we have,

a = 2, b = k and c = 3

As we are aware, Discriminant = b2 – 4ac

= (k)2 – 4(2) (3)

= k2 – 24

For the equal roots, we get,

Discriminant = 0

k2 – 24 = 0

k2 = 24

Therefore, k = ±√24 = ±2√6

(ii) kx(x – 2) + 6 = 0

or kx2 – 2kx + 6 = 0

On comparing the given equation with ax2 + bx + c = 0, we get

a = k, b = – 2k and c = 6

We are aware, Discriminant = b2 – 4ac

= ( – 2k)2 – 4 (k) (6)

= 4k2 – 24k

For the equal roots, we get,

b2 – 4ac = 0

4k2 – 24k = 0

4k (k – 6) = 0

It can be 4k = 0 or k = 6 = 0

k = 0 or k = 6

As a result, if k = 0, then the equation will not have the terms ‘x2 and ‘x.’

Hence, if this equation has two equal roots, k should be 6 only.

 

 

 

Question 21. Find out the natural number whose square diminished by 84 is equal to the thrice for 8 more than the given number.

Answer 21:

Assume the natural number = ‘x’.

As per the question,

We find an equation,

x² – 84 = 3(x+8)

x² – 84 = 3x + 24

x² – 3x – 84 – 24 = 0

x² – 3x – 108 = 0

x² – 12x + 9x – 108 = 0

x(x – 12) + 9(x – 12) = 0

(x + 9) (x – 12)

 x = -9 as well as x = 12

As natural numbers can’t be negative.

Thus, the number is 12.

 

 

 

 

Question 22. A natural number, when increased by 12, is the same as 160 times it is reciprocal. Find out the number.

Answer 22:

Assume the natural number = x

If the number increased by 12 = x + 12

Then,

Reciprocal of the number = 1/x

As per the question, we get,

x + 12 = 160 times of reciprocal for x

x + 12 = 160/ x

x( x + 12 ) = 160

x2 + 12x – 160 = 0

x2 + 20x – 8x – 160 = 0

x( x + 20) – 8( x + 20)= 0

(x + 20) (x – 8) = 0

x + 20 = 0 or x – 8 = 0

x = – 20 and x = 8

Natural numbers can’t be negative.

Hence,

The required number = x = 8

 

 

 

 

Question 23. If Zeba was younger by 5 years than what she really is, then the square for her age (in years) would have been 11 more than five times her actual age. What is her age now?

Answer 23:

Assume Zeba’s age = x

As per the question,

(x-5)²=11+5x

x²+25-10x=11+5x

x²-15x+14=0

x²-14x-x+14=0

x(x-14)-1(x-14)=0

x=1 or x=14

We have to ignore 1 like 5 years younger than 1 can’t happen.

Hence, Zeba’s present age = 14 years.

 

 

 

Question 24. Show the following situations in the form for quadratic equations:

(i) The area for a rectangular plot is 528 m2. The length for the plot (in metres) is one more than twice the breadth. We need to find out the length and breadth of the plot.

(ii) The product for two consecutive positive integers is 306. We need to find out the integers.

(iii) Rohan’s mother is 26 years older than him. The product for their ages (in years) 3 years from now will be 360. We would like to find out Rohan’s present age.

(iv) A train travels a distance of 480 km at a constant speed. If the speed had been 8 km/h less, what would be the speed of the train?

Answer 24:

(i) We can consider,

Breadth for the rectangular plot = x m

Therefore, the length for the plot = (2x + 1) m.

As we get,

Area for the rectangle = length × breadth = 528 m2

Substituting the value for the length and breadth for the plot in the formula, we find,

(2x + 1) × x = 528

2x2 + x =528

2x2 + x – 528 = 0

Hence, the length and breadth for the plot satisfy the quadratic equation, i.e. 2x2 + x – 528 = 0, which is the required representation for the problem mathematically.

(ii) We can consider,

The first integer number is = x

Therefore, the next consecutive positive integer can be = x + 1

Product for two consecutive integers = x × (x +1) = 306

 x2 + x = 306

x2 + x – 306 = 0

Hence, the two integers x and x+1, satisfy the quadratic equation, x2 + x – 306 = 0, which is the required representation for the problem mathematically.

(iii) We can consider,

Age for Rohan’s = x  years

Hence, according to the given question,

Rohan’s mother’s age is = x + 26

After the 3 years,

Age for Rohan’s = x + 3

Age for Rohan’s mother will be = x + 26 + 3 = x + 29

The product for their ages after 3 years will be same as the 360, so that

(x + 3)(x + 29) = 360

 x2 + 29x + 3x + 87 = 360

x2 + 32x + 87 – 360 = 0

x2 + 32x – 273 = 0

Hence, the age for Rohan and his mother satisfies the quadratic equation, I e. x2 + 32x – 273 = 0, which is the needed representation for the problem mathematically.

3

(iv) We can consider,

The speed of train = x  km/h

Or

Time taken for travelling 480 km = 480/x km/hr

According to the second condition, 

the speed of train = (x – 8) km/h

We also have

 The train will take 3 hours for covering r the same distance.

Hence,, time taken to travel 480 km = (480/x)+3 km/h

As we are aware,,

Speed × Time = Distance

Hence,

(x – 8)(480/x )+ 3 = 480

 480 + 3x – 3840/x – 24 = 480

3x – 3840/x = 24

x2 – 8x – 1280 = 0

Hence, the speed of the train satisfies the quadratic equation, I.e., x2 – 8x – 1280 = 0, which is the required representation for the problem mathematically.

 

 

 

Question 25. Find out two numbers whose sum is 27 and the product is 182.

Answer 25:

Assume the first number is x and the second number is 27 – x.

Thus, the product for two numbers

x(27 – x) = 182

x2 – 27x – 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) -14(x – 13) = 0

(x – 13)(x -14) = 0

Therefore, either, x – 13= 0  or x – 14 = 0

 x = 13 or x = 14

So, if first number = 13, then second number = 27 – 13 = 14

When the first number = 14, the second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

 

 

 

Question 26. Find out the two consecutive positive integers, sum of whose squares is 365.

Answer 26:

Assume the two consecutive positive integers as x and x + 1.

Hence,, according to the given question,

x2 + (x + 1)2 = 365

x2 + x2 + 1 + 2x = 365

 2x2 + 2x – 364 = 0

 x2 + x – 182 = 0

 x2 + 14x – 13x – 182 = 0

x(x + 14) -13(x + 14) = 0

(x + 14)(x – 13) = 0

Therefore, we get, x + 14 = 0 or x – 13 = 0,

 x = – 14 or x = 13

As the integers are positive, x could be 13 only.

We conclude,  x + 1 = 13 + 1 = 14

Thus, two consecutive positive integers will be 13 and 14.

 

 

 

Question 27. The altitude for a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find out the other two sides.

Answer 27:

Assume the base of the right triangle is x cm.

Given that, the altitude for right triangle = (x – 7) cm

From the Pythagoras theorem, we get

Base2 + Altitude2 = Hypotenuse2

So,

x2 + (x – 7)2 = 13×2

x2 + x2 + 49 – 14x = 169

 2x2 – 14x – 120 = 0

x2 – 7x – 60 = 0

x2 – 12x + 5x – 60 = 0

 x(x – 12) + 5(x – 12) = 0

 (x – 12)(x + 5) = 0

As a result, we can get x – 12 = 0 or x + 5 = 0,

 x = 12 or x = – 5

As the sides cannot be negative, x can only be 12.

Hence, the base for the given triangle is 12 cm and the altitude for this triangle will be (12 – 7) cm = 5 cm.

 

 

 

Question 28. The cottage industry produces a certain number of pottery articles in a day. It was observed on the particular day that the cost for production of each article (in rupees) was 3 more than twice the number of articles produced on that particular day. If the total cost of production on that day was Rs.90, find out the number of articles produced and the cost for each article.

Answer 28:

Assume the number of articles produced be x.

Hence, cost for production of each article = Rs (2x + 3)

Given that the total cost for production is Rs.90

Hence, 

x(2x + 3) = 90

 2x2 + 3x – 90 = 0

2x2 + 15x -12x – 90 = 0

x(2x + 15) -6(2x + 15) = 0

(2x + 15)(x – 6) = 0

As a result,  we can say 2x + 15 = 0 or x – 6 = 0

x = -15/2 or x = 6

Since the number for the articles produced can only be a positive integer, hence, x can only be 6.

Therefore, the number of articles produced = 6

Cost for each article = 2 × 6 + 3 = Rs 15.

 

 

 

 

Question 29. Find out the roots of the following quadratic equations, if they exist by completing the square method:

(i) 2x2 – 7x +3 = 0

(ii) 2x2 + x – 4 = 0

(iii) 4x2 + 4√3x + 3 = 0

(iv) 2x2 + x + 4 = 0

Answer 29:

(i) 2x2 – 7x + 3 = 0 

 2x2 – 7x = – 3

On dividing by 2 on both the sides, we find

 x2 -7x/2 = -3/2

 x2 -2 × x ×7/4 = -3/2

adding (7/4)2 to both the sides of the equation, we find

(x)2-2×x×7/4 +(7/4)2 = (7/4)2-3/2

(x-7/4)2 = (49/16) – (3/2)

(x-7/4)2 = 25/16

(x-7/4)2 = ±5/4

x = 7/4 ± 5/4

x = 7/4 + 5/4 or x = 7/4 – 5/4

x = 12/4 or x = 2/4

x = 3 or x = 1/2

(ii) 2x2 + x – 4 = 0

 2x2 + x = 4

On dividing both the sides of the equation by 2, we find

⇒ x2 +x/2 = 2

Now, adding (1/4)2 to both sides of the equation, we find

(x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2

 (x + 1/4)2 = 33/16

 x + 1/4 = ± √33/4

 x = ± √33/4 – 1/4

x = (± √33-1)/4

Thus, we have x = (√33-1)/4 or x = (-√33-1)/4

(iii) 4×2 + 4√3x + 3 = 0

On converting the equation into the a2+2ab+b2 form, we find,

(2x)2 + 2 × 2x × √3 + (√3)2 = 0

(2x + √3)2 = 0

(2x + √3) = 0 and (2x + √3) = 0

Thus, we have r x = -√3/2 or x = -√3/2.

(iv) 2x2 + x + 4 = 0

 2x2 + x = -4

On dividing both sides of the equation with 2, we find

 x2 + 1/2x = 2

x2 + 2 × x × 1/4 = -2

On adding (1/4)2 to both the sides of the equation, we find

(x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2

(x + 1/4)2 = 1/16 – 2

 (x + 1/4)2 = -31/16

Since, the square of numbers cannot be negative.

Thus there is no real root for the given equation, 2x2 + x + 4 = 0.

 

 

 

Question 30. Find out the roots for the quadratic equations given by applying the quadratic formula.

Answer 30:

(i) 2x2 – 7x + 3 = 0

compare the given equation by ax2 + bx + c = 0, we find,

a = 2, b = -7 and c = 3

Using the quadratic formula, we find

x = [-b ± √(b2 – 4ac)] /2a

 x = (7±√(49 – 24))/4

x = (7±√25)/4

 x = (7±5)/4

x = (7+5)/4 or x = (7-5)/4

x = 12/4 or 2/4

Hence,  x = 3 or 1/2

(ii) 2x2 + x – 4 = 0

 compare the given equation by ax2 + bx + c = 0, we find,

a = 2, b = 1 and c = -4

 Using quadratic formula, we find,,

x = [-b ± √(b2 – 4ac)] /2a

x = (-1±√1+32)/4

x = (-1±√33)/4

thus, x = (-1+√33)/4 or x = (-1-√33)/4

(iii) 4x2 + 4√3x + 3 = 0

Compare the given equation by ax2 + bx + c = 0, we find,

a = 4, b = 4√3 and c = 3

Using quadratic formula, we find,

x = [-b ± √(b2 – 4ac)] /2a

 x = (-4√3±√48-48)/8

x = (-4√3±0)/8

thus,  x = -√3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0

Compare the given equation by ax2 + bx + c = 0, we find,

a = 2, b = 1 and c = 4

Using quadratic formula, we find,

x = [-b ± √(b2 – 4ac)] /2a

 x = (-1±√1-32)/4

x = (-1±√-31)/4

As we now know, the square for a number can never be negative. 

Hence, there is no real solution for the given equation.

 

 

 

 

Question 31. Find the roots for the given equations:

(i) x-1/x = 3, x ≠ 0

(ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7

Answer 31:

(i) x-1/x = 3

⇒ x2 – 3x -1 = 0

Compare the given equation by ax2 + bx + c = 0, we find,

a = 1, b = -3 and c = -1

Using quadratic formula, we find,,

x = [-b ± √(b2 – 4ac)] /2a

x = (3±√9+4)/2

x = (3±√13)/2

So, x = (3+√13)/2 or x = (3-√13)/2

(ii) 1/x+4 – 1/x-7 = 11/30

 

x-7-x-4/(x+4)(x-7) = 11/30

-11/(x+4)(x-7) = 11/30

(x+4)(x-7) = -30

 x2 – 3x – 28 = 30

 x2 – 3x + 2 = 0

We can solve the equation by the factorisation method, we get,

 x2 – 2x – x + 2 = 0

 x(x – 2) – 1(x – 2) = 0

(x – 2)(x – 1) = 0

Thus,  x = 1 or 2

 

Question 32. A train travelling at a constant speed of 360 km will have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find out the original speed of the train.

Answer 32:

Let assume original speed of train = x km/h

We know,

Time = distance/speed

According to the question we have,

Time is taken by the train = 360/x hour

also, Time taken by the train its speed increase 5 km/h = 360/( x + 5)

It is given that,

Time taken by the train in first – time taken by the train for the 2nd case = 48 min = 48/60 hour

360/x – 360/(x +5) = 48/60 = 4/5

360(1/x – 1/(x +5)) = 4/5

360 ×5/4 (5/(x² +5x)) =1

450 × 5 = x² + 5x

x² +5x -2250 = 0

x = (-5± √ (25+9000))/2

= (-5 ±√ (9025) )/2

= (-5 ± 95)/2

= -50, 45

as x ≠ -50 because speed cannot be negative

thus, x = 45 km/h

therefore, original speed of train = 45 km/h

 

 

 

 

Question 33.(i) John and Jivanti together have 45 marbles.  If both of them lose 5 marbles each, and the product for the number of each marble  is now 124, then  find out how many marbles they had to begin with.

(ii) The cottage industry produces a certain number of toys in a day. The cost of production for each toy in rupees was found to be 55 minus the number for toys produced in a day. On a particular day, the total cost for production was  750. We would like to find out the number of toys produced on that particular day.

Answer 33:

(i) Assume the number for marbles John has = x.

Hence, the number of marble Jivanti have = 45 – x

After losing the 5 marbles each,

The number for marbles John have = x – 5

Number for marble Jivanti have = 45 – x – 5 = 40 – x

We can say the product for their marbles is 124.

Thus,

 (x – 5)(40 – x) = 124

 x2 – 45x + 324 = 0

 x2 – 36x – 9x + 324 = 0

x(x – 36) -9(x – 36) = 0

(x – 36)(x – 9) = 0

Therefore, we can say that

x – 36 = 0 or x – 9 = 0

 x = 36 or x = 9

Hence,

When John’s marbles = 36,

For, Jivanti’s marbles = 45 – 36 = 9

So, if John’s marbles = 9,

Then, we have Jivanti’s marbles = 45 – 9 = 36

(ii) Assume the number of toys produced in a day be x.

Hence, the cost of production of each toy = Rs(55 – x)

Given that, the total cost for production of the toys = Rs 750

Thus, x(55 – x) = 750

x2 – 55x + 750 = 0

x2 – 25x – 30x + 750 = 0

x(x – 25) -30(x – 25) = 0

 (x – 25)(x – 30) = 0

Therefore, if x -25 = 0 or x – 30 = 0

x = 25 or x = 30

Thus, the number of toys produced in a day will be either 25 or 30.

 

 

 

Question 34. The sum of the reciprocals of Rehman’s ages in years 3 years ago and 5 years from now is 1/3. Find out his present age.

Answer 34:

Assume the present age of Rahman is x years.

Three years from now, Rehman’s age was (x – 3) years.

After five years, his age will be (x + 5) years.

Given that, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

hence, 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

 3(2x + 2) = (x-3)(x+5)

6x + 6 =  x2 + 2x – 15

 x2 – 4x – 21 = 0

 x2 – 7x + 3x – 21 = 0

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0

And

x = 7, -3

As we are aware, age can’t be negative.

Hence, Rahman’s present age is 7 years.

 

Question 35. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product for their marks would have been 210. Find out her marks in the two subjects.

Answer 35:

Assume the marks of Shefali in Mathematics be x.

 The marks in English will be 30 – x.

as per given in the question,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

 -x2 + 25x + 54 = 210

 x2 – 25x + 156 = 0

 x2 – 12x – 13x + 156 = 0

x(x – 12) -13(x – 12) = 0

(x – 12)(x – 13) = 0

And

x = 12, 13

Hence, when the marks in Mathematics are 12, the marks in English would be 30 – 12 = 18 as well as the marks in Mathematics are 13, the marks in English will be 30 – 13 = 17.

 

Question 36. The diagonal for a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find out the sides of the field.

Answer 36:

Assume, the shorter side of the rectangle is x m.

The larger side of the rectangle = (x + 30) m

Diagonal of the rectangle = √[x2 + (x + 30)2]

As we have the length of the diagonal is = x + 30 m

Hence,

√[x2 + (x + 30)2] = x + 60

x2 + (x + 30)2 = (x + 60)2

x2 + x2 + 900 + 60x = x2 + 3600 + 120x

x2 – 60x – 2700 = 0

x2 – 90x + 30x – 2700 = 0

x(x – 90) + 30(x -90) = 0

(x – 90)(x + 30) = 0

And

x = 90, -30

As a result, the side of the field cannot be negative. Hence, the length of the shorter side will be 90 m and the length for the larger side will be (90 + 30) m = 120 m.

Question 37. The difference of squares for two numbers is 180. The square for the smaller number is 8 times the larger number. Find out the two numbers.

Answer 37:

Assume the larger and smaller number be x and y respectively.

As per the question,

x2 – y2 = 180 and y2 = 8x

x2 – 8x = 180

 x2 – 8x – 180 = 0

 x2 – 18x + 10x – 180 = 0

 x(x – 18) +10(x – 18) = 0

(x – 18)(x + 10) = 0

x = 18, -10

As a result, the larger number cannot be considered a negative number, 8 times for the larger number will be negative and therefore, the square of the smaller number will be negative which cannot be possible.

hence, the larger number will be 18 only.

x = 18

Thus, y2 = 8x = 8 × 18 = 144

 y = ±√144 = ±12

And

Smaller number = ±12

Hence, the numbers are 18 and 12 or 18 and -12.

Question 38. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken  1 hour less for the same journey. Find out the speed of the train.

Answer 38:

We have,

Distance = 360 km

Considering x as the speed, then the time taken will be

t = 360/x

When the speed is increased by 5 km/h the speed will be (x + 5) km/h

The distance will be the equal,

t = 360/(x + 5)

We get,

Time with the original speed – Time with the increased speed = 1

360/x – 360/(x + 5) = 1

LCM for= x (x + 5)

[360 (x + 5) – 360x]/x(x + 5) = 1

360 x + 1800 – 360x = x (x + 5)

x2 + 5x = 1800

x2 + 5x – 1800 = 0

x2 + 45x – 40x – 1800 = 0

x (x + 45) – 40 (x + 45) = 0

(x – 40) (x + 45) = 0

x = 40 km/hr

we know, the value of speed cannot be negative.

Hence, the speed of the train is 40 km/h.

Question 39. Two water taps together could fill a tank in (9+⅜) hours. The tap for larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find out the time in which each tap can separately fill the tank.

Answer 39:

Assume time taken by the smaller pipe to fill the tank = x hr.

The time taken by the larger pipe = (x – 10) hr

Part of the tank filled by the smaller pipe in 1 hour = 1/x

Part of the tank filled by the larger pipe in 1 hour = 1/(x – 10)

As we have, the tank could be filled in

(9+⅜)= 75/8 hours with both the pipes together.

Hence,

1/x + 1/x-10 = 8/75

x-10+x/x(x-10) = 8/75

 2x-10/x(x-10) = 8/75

 75(2x – 10) = 8x2 – 80x

150x – 750 = 8x2 – 80x

8x2 – 230x +750 = 0

8x2 – 200x – 30x + 750 = 0

 8x(x – 25) -30(x – 25) = 0

(x – 25)(8x -30) = 0

 x = 25, 30/8

The time taken by the smaller pipe cannot be 30/8 = 3.75 hours, and the time taken for the larger pipe will become negative, which is logically not possible.

So, the time taken by each of the smaller pipe as well as the larger pipe will be 25 and 25 – 10 =15 hours respectively.

Question 40. An express train takes 1 hour less than  a passenger train to travel 132 km from Mysore to Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find out the average speed of the two trains.

Answer 40:

Assume, the average speed of the passenger train =  x km/h.

Average speed of the express train = (x + 11) km/h

Given that, the time taken for the express train to cover 132 km is 1 hour less as compared to the passenger train to cover the equal distance. 

Hence,

(132/x) – (132/(x+11)) = 1

132(x+11-x)/(x(x+11)) = 1

132 × 11 /(x(x+11)) = 1

132 × 11 = x(x + 11)

x2 + 11x – 1452 = 0

x2 +  44x -33x -1452 = 0

 x(x + 44) -33(x + 44) = 0

(x + 44)(x – 33) = 0

And

x = – 44, 33

As we are aware, 

Speed cannot be negative.

Hence, the speed of the passenger train will be 33 km/h and so the speed of the express train will be 33 + 11 = 44 km/h.

Question 41. The sum of the areas of two squares is 468 m2. If the difference for their perimeters is 24 m, find out the sides of the two squares.

Answer 41:

Assume the sides of the two squares be x m and y m.

Hence, their perimeter will be 4x and 4y respectively

The area of the squares will be x2 and y2 

Given that,

4x – 4y = 24

x – y = 6

x = y + 6

And, x2 + y2 = 468

(6 + y2) + y2 = 468

36 + y2 + 12y + y2 = 468

2y2 + 12y + 432 = 0

y2 + 6y – 216 = 0

y2 + 18y – 12y – 216 = 0

y(y +18) -12(y + 18) = 0

(y + 18)(y – 12) = 0

And

 y = -18, 12

As we are aware, the side of a square cannot be negative.

So, the sides of the squares are 12 m and (12 + 6) m = 18 m.

 

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