Well-designed AP Board Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3 offers step-by-step explanations to help students understand problem-solving strategies.

## Quadratic Equations Class 10 Exercise 4.3 Solutions – 10th Class Maths 4.3 Exercise Solutions

Question 1.

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

i) 2x^{2} – 3x + 5 = 0

ii) 3x^{2} – 4\(\sqrt{3}\)x + 4 = 0

iii) 2x^{2} – 6x + 3 = 0

Solution:

i) 2x^{2} – 3x + 5 = 0

Given, quadratic equation is 2x^{2} – 3x + 5 = 0 is in the form of ax^{2} + bx + c = 0 a = 2, b = -3 and c = 5

Discriminant D = b^{2} – 4ac

= (-3)^{2} – 4(2)(3)

= 9 – 40 = -31 < 0

∴ D < 0

Therefore, roots of the given equation are not real.

ii) 3x^{2} – 4\(\sqrt{3}\)x + 4 = 0.

Given quadratic equation is 3x^{2} – 4\(\sqrt{3}\)x + 4 = 0 in the form of ax^{2} + bx + c = 0

a = 3, b = -4\(\sqrt{3}\) and c = 4

Discriminant D = b^{2} – 4ac

= (-4\(\sqrt{3}\))^{2} – 4(3)(4)

= 48 – 48 = 0

∴ D = 0

Therefore, roots of the given equation are real and equal.

iii) 2x^{2} – 6x + 3 = 0

Given quadratic equation is 2x^{2} – 6x + 3 = 0 is in the form of ax^{2} + bx + c = 0

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

Discriminant D = b^{2} – 4ac

= (-6)^{2} – 4(2)(3)

= 36 – 24 = +12 > 0

Therefore, roots of the given equation are real and distinct.

Question 2.

Find the values of k for each of the following quadratic equations, so that they have two equal roots.

i) 2x^{2} + kx + 3 = 0

ii) kx (x – 2) + 6 = 0

Solution:

i) 2x^{2} + kx + 3 = 0

Given quadratic equation 2x^{2} + kx + 3 = 0 is in the form of ax + bx + c = 0 and roots are a = 2, b = k and c = 3 real and equal.

Discriminant (D) = b^{2} – 4ac = 0

⇒ k^{2} – 4(2)(3) = 0

⇒ k^{2} – 24 = 0

⇒ k^{2} = \(\sqrt{24}\)

⇒ k^{2} = \(\sqrt{24}\)

∴ k = ±\(\sqrt{24}\)

ii) kx (x – 2) + 6 = 0

Given quadratic equation

kx (x – 2) + 6 = 0 = kx^{2} – 2kx + 6 = 0 is in the form of ax^{2} + bx + c = 0 and roots are real and equal.

a = k, b = -2k and c = 6

Discriminant (D) = b^{2} – 4ac

⇒ (-2k)^{2} – 4(k)(6) = 0

⇒ 4k^{2} – 24k = 0

⇒ 4k (k – 6) = 0

⇒ k = 0 (or) k = 6.

⇒ k = 6 only

Question 3.

Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m^{2}? If so, find its length and breadth.

Solution:

Let the breadth of a rectangular mango grove = x m.

Then the length of rectangle = 2x

Area of rectangle = l • b = 800 m^{2}

⇒ x • 2x = 800

⇒ 2x^{2} = 800

⇒ x^{2} = \(\frac{800}{2}\)

⇒ x^{2} = 400

⇒ x = \(\sqrt{400}\) = 20

∴ breadth of mango grove x = 20 m

∴ Length of mango grove = 2x = 2 × 20 = 40 m

∴ Length and breadth of grove are 40 m and 20 m.

Question 4.

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.

Solution:

Let the present ages of two friends be x years and (20 – x) years respectively. Four years ago their ages are (x – 4) years and (20 – x – 4) years.

Product = 48

⇒ (x – 4) (20 – x – 4) = 48 .

⇒ (x – 4) (16 – x) = 48

⇒ 16x – x^{2} – 64 + 4x = 48

⇒ 20x – x^{2} – 64 = 48

⇒ x^{2} – 20x + 48 + 64 = 0

⇒ x^{2} – 20x + 112 = 0

This is in the form of ax2 + bx + c = 0

a = 1, b = – 20, c = 112

D = b^{2} – 4ac

= (-20)^{2} – 4(1)(112)

D = 400 – 448 = -48 < 0

⇒ D < 0

So, above equation does not have real roots. Therefore, the given situation is not possible.

Question 5.

Is it possible to design a rectangular park of perimeter 80 m and area 400 m^{2}? If so, find its length and breadth.

Solution:

Let the length and breadth of a rectangular park are l and b respectively.

Perimeter = 2 (l + b) = 80 m

⇒ l + b = \(\frac{80}{2}\) = 40 m

b = 40 – l.

Area = l × b = 400

l(40 – 1) = 400

40l – l^{2} = 4oo

l^{2} – 40l + 400 = 0

It is in the form of ax^{2} + bx + c = 0

a = 1, b = -40, c = 400

Discriminant D = b^{2} – 4ac

= (-40)^{2} – 4(1)(400)

= 1600 – 16000 = 0

D = 0

Therefore, given equation have real and equal roots. So, given situation is possible.

l^{2} – 40l + 400 = 0

l^{2} – 20l – 20l + 400 = 0

l(l – 20) – 20 (l – 20) = 0

(l – 20)(l – 20) = 0

Length l = 20 m, b = 40 – l

= 40 – 20

= 20 m

breadth = 20 m.