Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.4 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.4

Question 1.

Find the nature of the roots of the following quadratic equations. If real roots exist, find them. (A.P June is)

i) 2x^{2} – 3x + 5 = 0

Solution:

Given : 2x^{2} – 3x + 5 = 0

a = 2; b = -3; c = 5

Discriminant = b^{2} – 4ac

b^{2} – 4ac = (-3)^{2} – 4(2)(5)

= 9 – 40

= -31 < 0

Roots are imaginary,

ii) 3x^{2} – 4 \(\sqrt{3}\)x + 4 = 0

Solution:

Given : 3x^{2} – 4\(\sqrt{3}\)x + 4 = 0

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

b^{2} – 4ac = (-\(\sqrt{3}\))^{2} – 4(3)(4)

= 48 – 48 = 0

∴ Roots are real and equal and they are

\(\frac{-b}{2 a}\), \(\frac{-b}{2 a}\)

= \(\frac{-(-4 \sqrt{3})}{2 \times 3}\) = \(\frac{4 \sqrt{3}}{6}\) = \(\frac{2}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)

iii) 2x^{2} – 6x + 30

Solution:

Given : 2x^{2} – 6x + 3 = 0

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

b^{2} – 4ac = (-6)^{2} – 4(2)(3)

= 36 – 24 = 12 > 0

The roots are real and distinct.

They are

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

Solution:

Given : 2x^{2} + kx + 3 = 0 has equal roots

∴ b^{2} – 4ac = 0

Here a = 2; b = k; c = 3

b^{2} – 4ac = (k)^{2} – 4(2) (3) = 0

⇒ k^{2} – 24 = 0

⇒ k^{2} = 24

⇒ k = \(\sqrt{24}\)

= ±2 \(\sqrt{6}\)

ii) kx(x – 2) + 6 = 0

Solution:

Given : kx(x – 2) + 6 = 0

kx^{2} – 2kx + 6 = 0

As this Q.E. has equal roots.

b^{2} – 4ac = 0

Here a = k; b = -2k; c = 6

∴ b^{2} – 4ac = (-2k)^{2} – 4(k) (6) = 0

⇒ 4k^{2} – 24 k = 0

⇒ 4k(k – 6) = 0

⇒ k = 0 (or) k – 6 = 0

⇒ k = 0 (or) 6

But k = 0 is trivial

∴ k = 6

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. (A.P. June 15)

Solution:

Let the breadth = x m

Then length = 2x m

Area = length × breadth

= x. (2x) = 2x^{2} m^{2}

By problem 2x^{2} = 800

⇒ x^{2} = 400

and x = \(\sqrt{400}\) = ± 20

∴ Breadth x = 20 m and

Length 2x = 2 × 20 = 40 m

Question 4.

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

Solution:

Let the age one of the two friends be x years.

Then the age of the other = 20 – x

Then, 4 years ago their ages would be (x – 4) and (20 – x – 4) = 16 – x

∴ Problem (x – 4) (16 – x) = 48

⇒ x(16 – x) – 4(16 – x) = 48

⇒ 16x – x^{2} – 64 + 4x = 48

⇒ x^{2} – 20x + 112 = 0

Here a = 1; b = – 20; c = 112

b^{2} – 4ac = (-20)^{2} – 4(1) (112)

= 400 – 448 = -48 < 0

Thus the roots are not real.

∴ The situation is not possible.

Question 5.

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

Solution:

Given : Perimeter of a rectangle 2(1l + b) = 80

⇒ l + b = \(\frac{80}{2}\) = 40 —– (1)

Area of the rectangle, l × b = 400

It possible, let us suppose that length of the rectangle = x m say

Then its breadth by equation (1) = 40 – x

By problem area = x. (40 – x) = 400

⇒ 40x – x^{2} = 400

⇒ x^{2} – 40x + 400 = 0

Here a = 1; b = – 40; c = 400

b^{2} – 4ac = (-40)^{2} – 4(1)(400)

= 1600 – 1600

= 0

∴ The roots are real and equal.

They are \(\frac{-b}{2 a}\), \(\frac{-b}{2 a}\)

i.e., \(\frac{-(-40)}{2 \times 1}\) = \(\frac{40}{2}\) = 20

∴ The dimensions are 20m, 20m.

(∴ The park is in square shape)