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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.14

Question 1.
Write each of the following expression in terms of α + β and αβ
(i) $$\frac{\alpha}{3 \beta}+\frac{\beta}{3 \alpha}$$ (ii) $$\frac{1}{\alpha^{2} \beta}+\frac{1}{\beta^{2} \alpha}$$ (iii) (3α – 1) (3β – 1)
(3α – 1) (3β – 1) = 9αc – 3α – 3β + 1
= 9αβ – 3(α + β) + 1 (iv) $$\frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha}$$ Question 2.
The roots of the equation 2x2 – 7x + 5 = 0 are a and p. Find the value of [without solving the equation]
$$\text { (i) } \frac{1}{\alpha}+\frac{1}{\beta}$$
α and α are the roots of the equation 2x2 – 7x + 5 = 0
α + β = $$\frac { 7 }{ 2 }$$ ; αβ = $$\frac { 5 }{ 2 }$$
(i) $$\frac{1}{\alpha}+\frac{1}{\beta}$$ = $$\frac{\beta+\alpha}{\alpha \beta}$$
= $$\frac { 7 }{ 2 }$$ + $$\frac { 5 }{ 2 }$$ = $$\frac { 7 }{ 2 }$$ × $$\frac { 2 }{ 5 }$$ = $$\frac { 7 }{ 5 }$$

(ii) $$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$$ = ($$\frac { 7 }{ 2 }$$)2 – 2 × $$\frac { 5 }{ 2 }$$ ÷ $$\frac { 5 }{ 2 }$$
= $$\frac { 49 }{ 4 }$$ – 5 ÷ $$\frac { 5 }{ 2 }$$ = $$\frac { 49-20 }{ 4 }$$ ÷ $$\frac { 5 }{ 2 }$$
= $$\frac { 29 }{ 4 }$$ × $$\frac { 2 }{ 5 }$$ = $$\frac { 29 }{ 10 }$$ (iii) $$\frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2}$$  Question 3.
The roots of the equation x2 + 6x – 4 = 0 are a, p. Find the quadratic equation whose roots are
(i) α2 and β2
α and β are the roots of x2 + 6x – 4 = 0
α + β = -6; αβ = -4

(i) Sum of the roots = α2 + β2
= (α + β)2 – 2αβ
= 36 – 2 – (4) = 36 + 8
= 44
Product of the roots = α2 + β2
= (αβ)2
= (-4)2
= 16
x2 – (sum of the roots) x + Product of the roots = 0
x2 – (44)x + 16 = 0
x2 – 44x + 16 = 0

(ii) $$\frac{2}{\alpha}$$ and $$\frac{2}{\beta}$$
Sum of the roots = $$\frac{2}{\alpha}$$ + $$\frac{2}{\beta}$$
= $$\frac{2 \beta+2 \alpha}{\alpha \beta}=\frac{2(\alpha+\beta)}{\alpha \beta}$$
= $$\frac{2(-6)}{-4}=\frac{-12}{-4}=3$$
Product of the roots = $$\frac{2}{\alpha} \times \frac{2}{\beta}=\frac{4}{\alpha \beta}$$
= $$\frac { 4 }{ -4 }$$ = -1
x2 – (sum of the roots) x + Product of the roots = 0
x2 – 3x – 1 = 0

(iii) α2β and β2α
Sum of the roots = α2β + β2α
= αβ (α + β)
= -4 (-6) = 24
Product of the roots = α2β × β2α
= α2β3 = (αβ)3
= (-4)3 = -64
x2 – (Sum of the roots) x + Product of the roots = 0
x2 – 24x – 64 = 0 Question 4.
If α, β are the roots of 7x2 + ax + 2 = 0 and if β – α = $$\frac { 13 }{ 7 }$$ Find the values of a.
α and β are the roots of 7x2 + ax + 2 = 0
α + β = $$\frac { -a }{ 7 }$$; αβ = $$\frac { 2 }{ 7 }$$
Given β – α = – $$\frac { 13 }{ 7 }$$ ⇒ α – β = $$\frac { 13 }{ 7 }$$
Squaring on both sides
(α – β)2 = ($$\frac { 13 }{ 7 }$$)2
α2 + β2 = 2αβ = $$\frac { 169 }{ 49 }$$
(- $$\frac { a }{ 7 }$$)2 -4($$\frac { 2 }{ 7 }$$) = $$\frac { 169 }{ 49 }$$ ⇒ $$\frac{a^{2}}{49}-\frac{8}{7}=\frac{169}{49}$$
$$\frac{a^{2}}{49}$$ = $$\frac { 225 }{ 49 }$$ ⇒ a2 = $$\frac{225 \times 49}{49}$$
a2 = 225 ⇒ a = ± $$\sqrt { 225 }$$ = ± 15
The value of a = 15 or – 15

Question 5.
If one root of the equation 2y2, – ay + 64 = 0 is twice the other then find the values of a.
Let the roots be α and 2α
Here a = 2, b = – a, c = 64
Sum of the roots = – $$\frac { b }{ a }$$
α + 2α = $$\frac { a }{ 2 }$$
3α = $$\frac { a }{ 2 }$$
a = 6α …….(1)
Product of the roots = $$\frac { c }{ a }$$
α × 2α = $$\frac { 64 }{ 2 }$$ = 2α2 = 32
α2 = $$\frac { 32 }{ 2 }$$ = 16
α = $$\sqrt { 16 }$$ = ± 4
Substitute the value of a in (1)
When α = 4
a = 6(4)
a = 24
The Value of a is 24 or -24
When α = -4
a = 6(-4)
a = -24 Question 6.
If one root of the equation 3x2 + kx + 81 = 0 (having real roots) is the square of the other then find k.
Let α and α2 be the root of the equation 3x2 + kx + 81
Here a = 3, b = k, c = 81
Sum of the roots = – $$\frac { b }{ a }$$ = – $$\frac { k }{ 3 }$$
α + α2 = –$$\frac { k }{ 3 }$$
3α + 3α2 = -k ……..(1)
Product of the roots = $$\frac { c }{ a }$$ = $$\frac { 81 }{ 3 }$$ = 27
α × α2 = 27
α3 = 27 ⇒ α3 = 33
α = 3
Substitute the value of α = 3 in (1)
3(3) + 3(3)2 = -k
9 + 27 = -k ⇒ 36 = – k
∴ k = -36
The value of k = -36