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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.13

Question 1.

Determine the nature of the roots for the following quadratic equations

(i) 15.x^{2} + 11.x + 2 = 0

(ii) x^{2} – x – 1 = 0

(iii) \(\sqrt{2} t^{2}\) – 3t + \(3 \sqrt{2}\) = 0

(iv) 9y^{2} – \(6 \sqrt{2} y\) + 2 = 0

(v) 9a^{2}b^{2}x^{2} – 24abcdx + 16c^{2}d^{2} = 0 a ≠ 0, b ≠ 0

Solution:

(i) 15x^{2} + 11x + 2 = 0 comparing with ax^{2} + bx + c = 0.

Here a = 15, 6 = 11, c = 2.

Δ = b^{2} – 4ac

= 11^{2} -4 × 15 × 2

= 121 – 120

= 1 > 1.

∴ The roots are real and unequal.

(ii) x^{2} – x – 1 = 0,

Here a = 1, b = -1, c = -1 .

Δ = b^{2} – 4ac

= (-1)^{2} – 4 × 1 × -1

= 1 + 4 = 5 > 0.

∴ The roots are real and unequal.

(iii) \(\sqrt{2} t^{2}\) – 3t + \(3 \sqrt{2}\) = 0

Here a = \(\sqrt{2}\), b = -3, c = \(3\sqrt{2}\)

Δ = b^{2} – 4ac

= (-3)^{2} – 4 × \(\sqrt{2}\) × \(3\sqrt{2}\)

= 9 – 24 = -15 < 0.

∴ The roots are not real.

(iv) 9y^{2} – \(6 \sqrt{2} y\) + 2 = 0

a = 9, b = \(6\sqrt{2}\) , c = 2

Δ = b^{2} – 4ac

= (\(6\sqrt{2}\))^{2} – 4 × 9 × 2

= 36 × 2 – 72

= 72 – 72 = 0

∴ The roots are real and equal.

(v) 9a^{2}b^{2}x^{2} – 24abcdx + 16c^{2}d^{2} = 0

Δ = b^{2} – 4ac

= (-24abcd)^{2} – 4 × 9a^{2}b^{2} × 16c^{2}d^{2}

= 576a^{2}b^{2}c^{2}d^{2} – 576a^{2}b^{2}c^{2}d^{2}

= 0

∴ The roots are real and equal.

Question 2.

Find the value(s) of ‘A’ for which the roots of the following equations are real and equal.

(i) (5k – 6)x^{2} + 2kx + 1 = 0

Answer:

Here a = 5k – 6 ; b = 2k and c = 1

Since the equation has real and equal roots ∆ = 0.

∴ b^{2} – 4ac = 0

(2k)^{2} – 4(5k – 6) (1) = 0

4k^{2} – 20k + 24 = 0

(÷ 4) ⇒ k^{2} – 5k + 6 = 0

(k – 3) (k – 2) = 0

k -3 = 0 or k – 2 = 0

k = 3 or k = 2

The value of k = 3 or 2

(ii) kx^{2} + (6k + 2)x + 16 = 0

Answer:

Here a = k, b = 6k + 2; c = 16

Since the equation has real and equal roots

∆ = 0

b^{2} – 4ac = 0

(6k + 2)^{2} – 4(k) (16) = 0

36k^{2} + 4 + 24k – 4(k) (16) = 0

36k^{2} – 40k + 4 = 0

(÷ by 4) ⇒ 9k^{2} – 10k + 1 = 0

9k^{2} – 9k – k + 1 = 0

9k(k – 1) – 1(k – 1) = 0

9k (k – 1) -1 (k – 1) = 0

(k – 1) (9k – 1) = 0

k – 1 or 9k – 1 = 0

k = 1 or k = \(\frac { 1 }{ 9 } \)

The value of k = 1 or \(\frac { 1 }{ 9 } \)

Question 3.

If the roots of (a – b)x^{2} + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.

Solution:

(a – b)x^{2} + (b – c)x + (c – a) = 0

A = (a – b), B = (b – c), C = (c – a)

Δ = b^{2} – 4ac = 0

⇒ (b – c)^{2} – 4(a – b)(c – a)

⇒ b^{2} – 2bc + c^{2} -4 (ac – bc – a^{2} + ab)

⇒ b^{2} – 2bc + c^{2} – 4ac + 4bc + 4a^{2} – 4ab = 0

⇒ 4a^{2} + b^{2} + c^{2} + 2bc – 4ac – 4ab = 0

⇒- (-2a + b + c)^{2} = 0 [∵ (a + b + c) = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca)]

⇒ 2a + b + c = 0

⇒ 2 a = b + c

∴ a, b, c are in A.P.

Question 4.

If a, b are real then show that the roots of the equation

(a – b)x^{2} – 6(a + b)x – 9(a – b) = 0 are real and unequal.

Answer:

(a – b)x^{2} – 6(a + b)x – 9(a – b) = 0

Here a = a – b ; b = – 6 (a + b); c = – 9 (a – b)

∆ = b^{2} – 4ac

= [- 6(a + b)]^{2} – 4(a – b)[-9(a – b)]

= 36(a + b)^{2} + 36(a – b)(a – b)

= 36 (a + b)^{2} + 36 (a – b)^{2}

= 36 [(a + b)^{2} + (a – b)^{2}]

The value is always greater than 0

∆ = 36 [(a + b)^{2} + (a – b)^{2}] > 0

∴ The roots are real and unequal.

Question 5.

If the roots of the equation (c^{2} – ab)x^{2} – 2(a^{2} – bc)x + b^{2} – ac = 0 are real and equal prove that either a = 0 (or) a^{3} + b^{3} + c^{3} = 3abc.

Solution:

(c^{2} – ab)x^{2} – 2(a^{2} – bc)x + (b^{2} – ac) – 0

Δ = B^{2} – 4AC = 0 (since the roots are real and equal)

⇒ 4(a^{2′} – bc)^{2} – 4 (c^{2} – ab)(b^{2} – ac) = 0

⇒ 4(a^{4} – 2a^{2}bc + b^{2}c^{2}) – 4(c^{2}b^{2} – ab^{3} – ac^{3} + a^{2}bc) = 0

⇒ 4a^{4} + 4b^{2}c^{2} – 8a^{2}bc – 4c^{2}b^{2} + 4ab^{3} + 4ac^{3} – 4a^{2}bc = 0

⇒ 4a^{4}+ 4ab^{3} + 4ac^{3} – 4a^{2}bc – 8a^{2}bc = 0

⇒ 4a [a^{3} + b^{3} + c^{3}] = 0 or a = 0

⇒ a = 0 or [a^{3} + b^{3} + c^{3} – 3abc] = 0

⇒ a^{3} + b^{3} + c^{3} – 3abc = 0

⇒ a^{3} + b^{3} + c^{3} = 3abc or a = 0

Hence proved.