Students can practice TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise to get the best methods of solving problems.

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

Question 1.

Some points are plotted on a plane. Each point is joined with all remaining points by line segments. Find the number of points if the number of line segments are 10.

Solution:

Number of distinct line segments that can be formed out of n-points = \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)

Given : No. of line segments \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) = 10

⇒ n^{2} – n = 20

⇒ n^{2} – n – 20 = 0

⇒ n^{2} – 5n + 4n – 20 = 0

⇒ n(n- 5) + 4(n – 5) = 0

⇒ (n – 5) (n + 4) = 0

⇒ n – 5 = 0 (or) n + 4 = 0

⇒ n = 5 (or) – 4

∴ n = 5 [n – can’t be negative]

Question 2.

A two digit number is such that the product of the digits is 8. When 18 is added to the number, they interchange their places. Determine the number.

Solution:

Let the digit in the units place = x

Let the digit in the tens place = y

∴ The number = 10y + x

By interchanging the digits the number becomes 10x + y

By problem (10x + y) – (10y + x) = 18

⇒ 9x + 9y = 18

⇒ 9(x – y) = 18

⇒ x – y = \(\frac{18}{9}\) = 2

⇒ y = x – 2

(i.e.) digit in the tens place = x – 2

digit in the units place = x

Product of the digits = (x – 2) x

By problem x^{2} – 2x = 8

x^{2} – 2x – 8 = 0

⇒ x^{2} – 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x – 4) (x + 2) = 0

⇒ x – 4 = 0 (or) x + 2 = 0

⇒ x = 4 (or) x = -2

x = 4 [ ∵ x can’t be negative]

∴ The number is 24.

Question 3.

A piece of wire 8m in length, cut into two pieces and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m^{2} ?

[Hint: x + y = 8, (\(\frac{x}{4}\))^{2} + (\(\frac{y}{4}\))^{2} = 2

⇒ (\(\frac{x}{4}\))^{2} + \(\left(\frac{8-x}{4}\right)\) = 2

Solution:

Let the length of the first piece = x m

Then length of the second piece = (8 – x) m

∴ Side of the 1^{st} square = \(\frac{x}{4}\) m and

Side of the second square = \(\frac{8-x}{4}\) m

Sum of the areas = 2 m^{2}

\(\left(\frac{x}{4}\right)^2\) + \(\left(\frac{8-x}{4}\right)^2\) = 2

⇒ x^{2} + 64 + x^{2} – 16x = 16 × 2 = 32

⇒ 2x^{2} – 16x + 64 = 32

⇒ 2x^{2} – 16x + 32 = 0

⇒ 2(x^{2} – 8x + 16) = 0

⇒ x^{2} – 4x – 4x + 16 = 0

⇒ x(x – 4) – 4(x – 4) = 0

⇒ (x – 4) (x – 4) = 0

∴ x = 4

The cut should be made at the centre making two equal pieces of length 4m, 4m.

Question 4.

Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job by himself ?

Solution:

Let the time taken by Vinay to complete the job = x days

Then the time taken by Praveen to complete the job = x + 5 days

Both worked for 6 days to complete a job.

∴ Total work done by them is

\(\frac{6}{x}\) + \(\frac{6}{x+5}\) = 1

⇒ 6(2x + 5) = x^{2} + 5x

⇒ x^{2} – 7x – 30 = 0

⇒ x^{2} – 10x + 3x – 30 = 0

⇒ x(x – 10) + 3(x – 10) = 0

⇒ x – 10 = 0 (or) x + 3 = 0

⇒ x = 10(or) x = -3

x = 10 (∵ x can’t be negative)

∴ Time taken by Vinay = x = 10 days

Time taken by Praveen = x + 5 = 15 days.

Question 5.

Show that the sum of roots of a quadratic equation is \(\frac{-\mathbf{b}}{\mathbf{a}}\).

Solution:

Let the Q.E. = ax^{2} + bx + c = 0 (a ≠ 0)

⇒ ax^{2} + bx = -c

⇒ x^{2} + \(\frac{\mathrm{b}}{\mathrm{a}}\)x = \(\frac{-\mathrm{c}}{\mathrm{a}}\)

⇒ x^{2} + 2.\(x \cdot \frac{b}{2 a}\) = \(-\frac{c}{a}\)

sum of the roots

∴ Sum of roots of a Q.E. is \(\frac{-b}{a}\)

Question 6.

Show that the product of the roots of a Q.E is \(\frac{\mathbf{c}}{\mathrm{a}}\).

Solution:

Let the Q.E. = ax^{2} + bx + c = 0(a ≠ 0)

⇒ ax^{2} + bx = -c

Product of the roots

Question 7.

The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\), find the fraction.

Solution:

Let the numerator = x

then denominator = 2x + 1

Then the fraction = \(\frac{x}{2 x+1}\)

Its reciprocal = \(\frac{2 x+1}{x}\)

⇒ 105x^{2} + 84x + 21 = 116x^{2} + 58x

⇒ 11x^{2} – 26x – 21 = 0

⇒ 11x^{2} – 33x + 7x – 21 = 0

⇒ 11x (x – 3) + 7(x – 3) = 0

⇒ (x – 3)(11x + 7) = 0

⇒ x – 3 = 0 (or) 11x + 3 = 0

⇒ x = 3 (or) \(\frac{-3}{11}\)

∴x = 3

Numerator = 3;

Denominator = 2 × 3 + 1 = 7

Fraction = \(\frac{3}{7}\)