Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

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) 5/3, 4
(iii) −3/2, -1
(iv) -(2 – a)², (a + 5)²
Solution:
If the roots are given, general form of the quadratic equation is x² – (sum of the roots) x + product of the roots = 0.
(i) Sum of the roots = -9
Product of the roots = 20
The equation = x² – (-9x) + 20 = 0
⇒ x² + 9x + 20 = 0

(ii) Sum of the roots = 5/3
Product of the roots = 4
Required equation = x² – (sum of the roots)x + product of the roots
= 0
⇒ x² – 5/3x + 4 = 0
⇒ 3x² – 5x + 12 = 0

(iii) Sum of the roots = (−3/2)
(α + β) = −3/2
Product of the roots (αβ) = (-1)
Required equation = x² – (α + β)x + αβ = 0
x² – (−3/2)x – 1 = 0
2x² + 3x – 2 = 0

(iv) α + β = – (2 – a)²
αβ = (a + 5)²
Required equation = x² – (α + β)x – αβ = 0
⇒ x² – (-(2 – a)²)x + (a + 5)² = 0
⇒ x² + (2 – a)²x + (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) x² + 3x – 28 = 0
(ii) x² + 3x = 0
(iii) 3 + 1/a=10/a2
(iv) 3y² – y – 4 = 0

(i) x² + 3x – 28 = 0
Answer:
Sum of the roots (α + β) = -3
Product of the roots (α β) = -28

(ii) x² + 3x = 0
Answer:
Sum of the roots (α + β) = -3
Product of the roots (α β) = 0

(iii) 3 + 1/a=10/a2
3a² + a = 10
3a² + a – 10 = 0 comparing this with x² – (α + β)
x + αβ = 0