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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) 15x^{2} + 11x + 2 = 0

Answer:

Here a = 15, b = 11, c = 2

∆ = b^{2} – 4ac

∆ = 11^{2} – 4(15) × 2

= 121 – 120

∆ = 1 > 0

So the equation will have real and unequal roots

(ii) x^{2} – x – 1 = 0

Answer:

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

∆ = b^{2} – 4ac

= (-1)^{2} – 4(1)(-1)

= 1 + 4 = 5

∆ = 1 > 0

So the equation will have real and unequal roots.

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

Answer:

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

∆ = b^{2} – 4ac

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

= 9 – 24 = -15

∆ = -15 < 0

So the equation will have no real roots.

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

Answer:

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

∆ = b^{2} – 4ac

= (-6\(\sqrt { 2 }\))^{2} – 4(9) (2)

= 72 – 72

= 0

So the equation will have real and equal roots.

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

Answer:

Here a = 9a^{2}b^{2}; b = -24 abed, c = 16c^{2}d^{2}

∆ = 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

So the equation will have real and equal roots.

Question 2.

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

(i) (5k – 6)^{2} + 2kx + 1 = 0

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

Solution:

(5k – 6)x^{2} + 2kx + 1 :

a = (5k – 6), b = 2, c = 1

Δ = b^{2} – 4ac

⇒ (2k)^{2} – 4 (5k – 6)(1)

⇒ 4k^{2} – 20k + 24 = 0 [∵ Since the roots are real and equal)

⇒ k^{2} – 5k + 6 = 0

⇒ (k – 3)(k – 2) = 0

k = 3, 2

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

a = k, b = (6k + 2), c = 16

Δ = b^{2} – 4ac [∵ the roots are real and equal)

⇒ (6k + 2)^{2} – 4 × k × 16 = 0

⇒ 36k^{2} + 24k + 4 – 64k = 0

⇒ 36k^{2} – 40k + 4 =0

⇒ 36k^{2} – 36k – 4k + 4 =0

⇒ 36k (k – 1) – 4 (k – 1) = 0

⇒ 4 (k – 1) (9k – 1) =0

⇒ k = 1 or k = \(\frac{1}{9}\)

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.

Answer:

(a – b) x2 + (b – c) x + (c – a) = 0

Here a = (a – b);b = b – c ; c = c – a

Since the equation has real and equal roots ∆ = 0

∴ b^{2} – 4ac = 0

(b – c)^{2} – 4(a – b)(c – a) = 0

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

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

b^{2} + c^{2} + 2bc -4a (b + c) + 4a^{2} = 0

(b + c)^{2} – 4a (b + c) + 4a^{2} = 0

[(b+c) – 2a]^{2} = 0 [using a^{2} – 2ab + b^{2} = (a – b)^{2}]

b + c – 2a = 0

b + c = 2a

b + c = a + a

c – a = a – b (t_{2} – t_{1} = t_{3} – t_{2})

b,a,c are in A.P.

Question 4.

If a, b are real then show that the roots of the equation (a – b)^{2} – 6(a + b)x – 9(a – b) = 0 are real and unequal.

Solution:

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

Δ = b^{2} – 4ac

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

= 36(a + b)^{2} + 36(a – b)^{2}

= 36 (a^{2} + 2ab + b^{2}) + 36(a^{2} – 2ab + b^{2})

= 72a^{2} + 12b^{2}

= 72(a^{2} + 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

Answer:

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

Here a = c^{2} – ab ; b = – 2 (a^{2} – bc); c = b^{2} – ac

Since the roots are real and equal

∆ = b^{2} – 4ac

[-2 (a^{2} – bc)]^{2} – 4(c^{2} – ab) (b^{2} – ac) = 0

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

Divided by 4 we get

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

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

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

= a(a^{3} + b^{3} + c^{3}) = 3a^{2}bc

a^{3} + b^{3} + c^{3} = \(\frac{3 a^{2} b c}{a}\)

a^{3} + b^{3} + c^{3} = 3 abc

Hence it is proved