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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) \(\frac{\alpha}{3 \beta}+\frac{\beta}{3 \alpha}\)

Answer:

(ii) \(\frac{1}{\alpha^{2} \beta}+\frac{1}{\beta^{2} \alpha}\)

Answer:

(iii) (3α – 1) (3β – 1)

Answer:

(3α – 1) (3β – 1) = 9αc – 3α – 3β + 1

= 9αβ – 3(α + β) + 1

(iv) \(\frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha}\)

Answer:

Question 2.

The roots of the equation 2x^{2} – 7x + 5 = 0 are a and p. Find the value of [without solving the equation]

\(\text { (i) } \frac{1}{\alpha}+\frac{1}{\beta}\)

Answer:

α and α are the roots of the equation 2x^{2} – 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) \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)

Answer

= (\(\frac { 7 }{ 2 } \))^{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) \(\frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2}\)

Answer:

Question 3.

The roots of the equation x^{2} + 6x – 4 = 0 are a, p. Find the quadratic equation whose roots are

(i) α^{2} and β^{2}

Answer:

α and β are the roots of x^{2} + 6x – 4 = 0

α + β = -6; αβ = -4

(i) Sum of the roots = α^{2} + β^{2}

= (α + β)^{2} – 2αβ

= 36 – 2 – (4) = 36 + 8

= 44

Product of the roots = α^{2} + β^{2}

= (αβ)^{2}

= (-4)^{2}

= 16

The Quadratic equation is

x^{2} – (sum of the roots) x + Product of the roots = 0

x^{2} – (44)x + 16 = 0

x^{2} – 44x + 16 = 0

(ii) \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\)

Answer:

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

The Quadratic equation is

x^{2} – (sum of the roots) x + Product of the roots = 0

x^{2} – 3x – 1 = 0

(iii) α^{2}β and β^{2}α

Answer:

Sum of the roots = α^{2}β + β^{2}α

= αβ (α + β)

= -4 (-6) = 24

Product of the roots = α^{2}β × β^{2}α

= α^{2}β^{3} = (αβ)^{3}

= (-4)^{3} = -64

The Quadratic equation is

x^{2} – (Sum of the roots) x + Product of the roots = 0

x^{2} – 24x – 64 = 0

Question 4.

If α, β are the roots of 7x^{2} + ax + 2 = 0 and if β – α = \(\frac { 13 }{ 7 } \) Find the values of a.

Answer:

α and β are the roots of 7x^{2} + 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}

α^{2} + β^{2} = 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 = \(\frac{225 \times 49}{49}\)

a^{2} = 225 ⇒ a = ± \(\sqrt { 225 }\) = ± 15

The value of a = 15 or – 15

Question 5.

If one root of the equation 2y^{2}, – ay + 64 = 0 is twice the other then find the values of a.

Answer:

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 = \(\frac { 64 }{ 2 } \) = 2α^{2} = 32

α^{2} = \(\frac { 32 }{ 2 } \) = 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 3x^{2} + kx + 81 = 0 (having real roots) is the square of the other then find k.

Answer:

Let α and α^{2} be the root of the equation 3x^{2} + kx + 81

Here a = 3, b = k, c = 81

Sum of the roots = – \(\frac { b }{ a } \) = – \(\frac { k }{ 3 } \)

α + α^{2} = –\(\frac { k }{ 3 } \)

3α + 3α^{2} = -k ……..(1)

Product of the roots = \(\frac { c }{ a } \) = \(\frac { 81 }{ 3 } \) = 27

α × α^{2} = 27

α^{3} = 27 ⇒ α^{3} = 3^{3}

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