# NCERT Class 10 Math Chapter 4 Important Questions Answer – Quadratic Equations

## Class 10 Math Chapter 4 Important Question Answer

Q1. Find the root of the quadratic equation by factorization method.

Ans.

=

=

So, the roots of the equation are the values of x for which

= 0

Now,

x =

and

x =

Therefore, the roots of are .

Q2. For what value of k the roots of the quadratic equation 2x2 + kx + 3 = 0 are equal?

Ans. Quadratic equation has eqal roots when discriminant is equal to zero.

Here a = 2, b = k, c = 3.

Therefore, Discriminant b2 – 4ac = k2 – 4×2×3 = k2 – 24

Comparing it with 0,

k2 – 24 = 0

k =

Required values of k are -2√6 and 2√6.

Q3. Sum of the roots of the quadratic polynomnial 7x2 – 3x + 1 is __________.

Ans. a =  7, b = -3

Sum of roots =

Q4. Factorise the quadratic equation 2x2 + x – 6 = 0 into linear factors.

Ans. We have,

2x2 + x – 6 = 0

2x2 + 4x – 3x  – 6 = 0

2x(x + 2) -3(x + 2) = 0

(2x – 3)(x + 2) = 0

Therefore, required linear factors of equation 2x2 + x – 6 are (2x – 3)(x + 2).

Q5. The roots of the quadratic equation 6x2 – x – 2 = 0 are __________.

Ans. We have

6x2 – x – 2  = 0

6x2 – 4x + 3x – 2 = 0

2x(3x – 2) + 1(3x – 2) = 0

(2x + 1)(3x – 2) = 0

On comparing

2x + 1 = 0      and    3x – 2 = 0

x =    and   x =

Therefore, the roots of 6x2 – x – 2 = 0 are and .

Q6. For what values of K, quadratic equation x2 + Kx + 4 = 0 has equal roots?

Ans. Quadratic equation has eqal roots when discriminant is equal to zero.

Here a = 1, b = k, c = 4.

Therefore, Discriminant b2 – 4ac = k2 – 4×1×4 = k2 – 16

Comparing it with 0,

k2 – 16 = 0

k = ±4

Required values of k are -4 and 4.

Q7. Find discriminant of quadratic equation x2 + 7x – 60 = 0.

Ans. Here a = 1, b = 7, c = -60.

Therefore, discriminant b2 – 4ac = 72 – 4×1×(-60) = 49 + 240 = 289

Q8. Semi perimeter of a rectangular garden, whose length is 4m more: than its width is 36m. Find the dimensions of the garden.

Ans. let breadth of rectangle = x m.

Then length will be (x + 4) m.

According to question,

Semi perimeter of rectangle = length + breadth = x + 4 + x = 36

2x + 4 = 36

2x = 32

x = 16 m.

length = x + 4 = 16 + 4 = 20 m.

Therefore, required dimensions of garden are 20 m and 16 m.

Q9. Find the nature of roots of the following quadratic equation. If real root exist, then solve it: 2x2 – 6x + 3 = 0

Ans. Nature of roots are determined by the discriminant of quadratic equation.

Here a = 2, b = -6, c = 3.

Therefore, discriminant b2 – 4ac = (-6)2 – 4×2×3 = 36 – 24 = 12.

Hence, b2 – 4ac > 0, two distinct real roots exist.

=

Therefore required roots of given quadratic equation are .

Q10. Find two consecutive positive integers the sum of whose squares is 365.

Ans. Let the consecutive positive integers be x and x + 1.

According to question,

x2  + (x + 1)2 = 365

x2 + x2 + 1 + 2x = 365

2x2 + 2x – 364 = 0

Dividing by 2, we get

x2 + x – 182 = 0

x2 + 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x – 13)(x + 14) = 0

On comparing, we get

x – 13 = 0 i.e x = 13

x + 14 = 0 i.e x = -14

As we have to find positive integers. Therfore, x = -14 is not possible.

Therefore, required two consecutive positive numbers are 13 and 14.

Q11. Find two numbers whose sum is 37 and product is 300.

Ans. Let 1st number be x.

Then 2nd number will be (37 – x)

According to question,

x(37 – x) = 300

37x – x2 = 300

37x – x2 – 300 = 0

x2 – 37x + 300 = 0

x2 – 12x – 25x + 300 = 0

x(x – 12) -25(x – 12) = 0

(x – 25)(x – 12) = 0

x – 25 = 0                x – 12 = 0

x = 25                      x = 12

If 1st number is 25, then 2nd number will be 12.

Similarly, if 1st number is 12, then 2nd number will be 25.

Therefore, two pairs (12, 25) and (25, 12) are made such that two numbers whose sum is 37 and product is 300.

Q12. Find two numbers whose sum is 27 and product is 182.

Ans. Let 1st number be x.

Then 2nd number will be (27 – x)

According to question,

x(27 – x) = 182

27x – x2 = 182

27x – x2 – 300 = 0

x2 – 27x + 182= 0

x2 – 13x –14x + 182 = 0

x(x – 13) -14(x – 13) = 0

(x – 14)(x – 13) = 0

x – 14 = 0                x – 13 = 0

x = 14                      x = 13

If 1st number is 14, then 2nd number will be 13.

Similarly, if 1st number is 13, then 2nd number will be 14.

Therefore, two pairs (13, 14) and (14, 13) are made such that two numbers whose sum is 27 and product is 182.

Q13. Is it possible to design a rectangular mango grove whose length is twice its breadth and area is 800 m2? If so, find its length and breadth.

Ans. Let the breadth of the mango grove be x.

The length of the mango grove will be 2x.

Area of the mango grove = (2x) (x)= 2x2

2x= 80

x= 800/2 = 400

x2  – 400 =0

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

a = 1, b = 0, c = 400

As we know, discriminant = b2 – 4ac

=> (0)2 – 4 × (1) × ( – 400) = 1600

Here, b2 – 4ac > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

x = ±20

As we know, the value of length cannot be negative.

Therefore, the breadth of the mango grove = 20 m.

Length of the mango grove = 2 × 20 = 40 m

Q14. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Ans. Let’s the present  age of one friend is x years.

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of first friend = (x – 4) years

Age of second friend = (20 – x – 4) = (16 – x) years

As per the given question, we can write,

(x – 4) (16 – x) = 48

16x – x2 – 64 + 4x = 48

– x2 + 20x – 112 = 0

x2 – 20x + 112 = 0

Comparing the equation with ax2 + bx c = 0, we get

a = 1b = -20 and c = 112

Discriminant = b2 – 4ac

= (-20)2 – 4 × 112

= 400 – 448 = -48

b2 – 4ac < 0

Therefore, there will be no real solution possible for the equations. Hence, the condition doesn’t exist .

Q15. One side of a rectangle exceeds its other side by 2cm. If its area is 195 cm2 , then determine the sides of the rectangle.

Ans. Let breadth of rectangle be x cm.

Then, length of triangle will be x + 2 cm.

According to question,

Area of rectangle = x(x + 2) = 195

x2 + 2x = 195

x2 + 2x – 195 = 0

x2 + 15x – 13x – 195 = 0

x(x + 15) – 13(x + 15) = 0

(x – 13)(x + 15) = 0

x – 13 = 0    or      x + 15 = 0

x = 13                        x = -15

As side of rectangle can’t be negative.

Therefore, x = 13 is true.

So, required sides of rectangle are 13 cm and 15 cm.

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