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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

**10th Maths Exercise 3.9 Samacheer Kalvi Question 1.**

Determine the quadratic equations, whose sum and product of roots are

(i) -9, 20

(ii) \(\frac{5}{3}\), 4

(iii) \(\frac{-3}{2}\), -1

(iv) -(2 – a)^{2}, (a + 5)^{2}

Solution:

If the roots are given, general form of the quadratic equation is x^{2} – (sum of the roots) x + product of the roots = 0.

(i) Sum of the roots = -9

Product of the roots = 20

The equation = x^{2} – (-9x) + 20 = 0

⇒ x^{2} + 9x + 20 = 0

(ii) Sum of the roots = \(\frac{5}{3}\)

Product of the roots = 4

Required equation = x^{2} – (sum of the roots)x + product of the roots

= 0

⇒ x^{2} – \(\frac{5}{3}\)x + 4 = 0

⇒ 3x^{2} – 5x + 12 = 0

(iii) Sum of the roots = (\(\frac{-3}{2}\))

(α + β) = \(\frac{-3}{2}\)

Product of the roots (αβ) = (-1)

Required equation = x^{2} – (α + β)x + αβ = 0

x^{2} – (\(\frac{-3}{2}\))x – 1 = 0

2x^{2} + 3x – 2 = 0

(iv) α + β = – (2 – a)^{2}

αβ = (a + 5)^{2}

Required equation = x^{2} – (α + β)x – αβ = 0

⇒ x^{2} – (-(2 – a)^{2})x + (a + 5)^{2} = 0

⇒ x^{2} + (2 – a)^{2}x + (a + 5)^{2} = 0

**Ex 3.9 Class 10 Samacheer Question 2.**

Find the sum and product of the roots for each of the following quadratic equations

(i) x^{2} + 3x – 28 = 0

(ii) x^{2} + 3x = 0

(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)

(iv) 3y^{2} – y – 4 = 0

(i) x^{2} + 3x – 28 = 0

Answer:

Sum of the roots (α + β) = -3

Product of the roots (α β) = -28

(ii) x^{2} + 3x = 0

Answer:

Sum of the roots (α + β) = -3

Product of the roots (α β) = 0

(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)

3a^{2} + a = 10

3a^{2} + a – 10 = 0 comparing this with x^{2} – (α + β)

x + αβ = 0