# Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.9

Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## 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
Answer:
Sum of the roots = -9 and Product of the roots = 20
The Quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (-9) x + 20 = 0 ⇒ x2 + 9x + 20 = 0

(ii) $$\frac { 5 }{ 3 }$$, 4
Answer:
Sum of the roots = $$\frac { 5 }{ 3 }$$; Product of the roots = 4
The Quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – ($$\frac { 5 }{ 3 }$$) x + 4 = 0 ⇒ x2 – $$\frac { 5 }{ 3 }$$ x + 4 = 0
3x2 – 5x + 12 = 0

(iii) $$\frac { -3 }{ 2 }$$, -1
Answer:
Sum of the roots = $$\frac { -3 }{ 2 }$$; Product of the roots = -1
The Quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (-$$\frac { 3 }{ 2 }$$) x + (-1) = 0 ⇒ x2 + $$\frac { 3 }{ 2 }$$ x – 1 = 0
2x2 + 3x – 2 = 0

(iv) – (2 – a)2, (a + 5)2
Answer:
Sum of the roots = – (2 – a)2; Product of the roots = (a + 5)2
x2 – (sum of the roots) x + product of the roots = 0
x2 – [-(2 – a)2] x + (a + 5)2 = 0
x2 + (2 – a)2 x + (a + 5)2 = 0

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
Solution:
(i) x2 – (-3)x + (-28) = 0.
Comparing this with x2 – (α + β)x + αβ = 0.
(α + β) = Sum of the roots = -3
αβ = product of the roots = -28

(ii) x2 + 3x = 0 = x2 – (-3)x + 0 = 0
x2 – (α + β)x + αβ = 0
Sum of the roots α + β = -3
Products of the roots αβ =0

(iii) 3 + $$\frac { 1 }{ a }$$ = $$\frac{10}{a^{2}}$$
Answer:
Multiply by a2
3a2 + a = 10
3a2 + a – 10 = 0
Sum of the roots (α + β) = $$\frac { -1 }{ 3 }$$
Product of the roots (α β) = $$\frac { -10 }{ 3 }$$

(iv) 3y2 – y – 4 = 0
Answer:
Sum of the roots (α + β) = $$\frac { -(-1) }{ 3 }$$ = $$\frac { 1 }{ 3 }$$
Product of the roots (α β) = $$\frac { -4 }{ 3 }$$

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