Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

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 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 = \(\frac{5}{3}\)
Product of the roots = 4
Required equation = x2 – (sum of the roots)x + product of the roots
= 0
⇒ x2 – \(\frac{5}{3}\)x + 4 = 0
⇒ 3x2 – 5x + 12 = 0

(iii) Sum of the roots = (\(\frac{-3}{2}\))
(α + β) = \(\frac{-3}{2}\)
Product of the roots (αβ) = (-1)
Required equation = x2 – (α + β)x + αβ = 0
x2 – (\(\frac{-3}{2}\))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

Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

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 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
(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 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
3a2 + a = 10
3a2 + a – 10 = 0 comparing this with x2 – (α + β)
x + αβ = 0
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.9 1

Leave a Comment

Your email address will not be published. Required fields are marked *