10th Maths Book Back Question and Answers – Chapter 3 Exercise 3.13:

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Samacheer Kalvi 10th Maths Book Back Answers – Ex 3.13 Algebra

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Chapter 3

Exercise 3.13 Algebra

1. Determine the nature of the roots for the following quadratic equations
(i) 15.x² + 11.x + 2 = 0
(ii) x² – x – 1 = 0
(iii) 2–√t2 – 3t + 32–√ = 0
(iv) 9y² – 62–√y + 2 = 0
(v) 9a²b²x² – 24abcdx + 16c²d² = 0 a ≠ 0, b ≠ 0
Solution:
(i) 15x² + 11x + 2 = 0 comparing with ax² + bx + c = 0.
Here a = 15, 6 = 11, c = 2.
Δ = b² – 4ac
= 11² -4 × 15 × 2
= 121 – 120
= 1 > 1.
∴ The roots are real and unequal.

(ii) x² – x – 1 = 0,
Here a = 1, b = -1, c = -1 .
Δ = b² – 4ac
= (-1)² – 4 × 1 × -1
= 1 + 4 = 5 > 0.
∴ The roots are real and unequal.

(iii) 2–√t2 – 3t + 32–√ = 0
Here a = 2–√, b = -3, c = 32–√
Δ = b² – 4ac
= (-3)² – 4 × 2–√ × 32–√
= 9 – 24 = -15 < 0.
∴ The roots are not real.

(iv) 9y² – 62–√y + 2 = 0
a = 9, b = 62–√ , c = 2
Δ = b² – 4ac
= (62–√)² – 4 × 9 × 2
= 36 × 2 – 72
= 72 – 72 = 0
∴ The roots are real and equal.

(v) 9a²b²x² – 24abcdx + 16c²d² = 0
Δ = b² – 4ac
= (-24abcd)² – 4 × 9a²b² × 16c²d²
= 576a²b²c²d² – 576a²b²c²d²
= 0
∴ The roots are real and equal.

2. Find the value(s) of ‘A’ for which the roots of the following equations are real and equal.
(i) (5k – 6)x² + 2kx + 1 = 0
Answer:
Here a = 5k – 6 ; b = 2k and c = 1
Since the equation has real and equal roots ∆ = 0.

∴ b² – 4ac = 0
(2k)² – 4(5k – 6) (1) = 0
4k² – 20k + 24 = 0
(÷ 4) ⇒ k² – 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) kx² + (6k + 2)x + 16 = 0
Answer:
Here a = k, b = 6k + 2; c = 16
Since the equation has real and equal roots

∆ = 0
b² – 4ac = 0
(6k + 2)² – 4(k) (16) = 0
36k² + 4 + 24k – 4(k) (16) = 0
36k² – 40k + 4 = 0
(÷ by 4) ⇒ 9k² – 10k + 1 = 0
9k² – 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 = 19
The value of k = 1 or 19

3. If the roots of (a – b)x² + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.
Solution:
(a – b)x² + (b – c)x + (c – a) = 0
A = (a – b), B = (b – c), C = (c – a)
Δ = b² – 4ac = 0
⇒ (b – c)² – 4(a – b)(c – a)
⇒ b² – 2bc + c² -4 (ac – bc – a² + ab)
⇒ b² – 2bc + c² – 4ac + 4bc + 4a² – 4ab = 0
⇒ 4a² + b² + c² + 2bc – 4ac – 4ab = 0
⇒- (-2a + b + c)² = 0 [∵ (a + b + c) = a² + b² + c² + 2ab + 2bc + 2ca)]
⇒ 2a + b + c = 0
⇒ 2 a = b + c
∴ a, b, c are in A.P.

4. If a, and b are real then show that the roots of the equation
(a – b)x² – 6(a + b)x – 9(a – b) = 0 are real and unequal.
Answer:
(a – b)x² – 6(a + b)x – 9(a – b) = 0
Here a = a – b ; b = – 6 (a + b); c = – 9 (a – b)
∆ = b² – 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.

5.If the roots of the equation (c² – ab)x² – 2(a² – bc)x + b² – ac = 0 are real and equal prove that either a = 0 (or) a³ + b³ + c³ = 3abc.
Solution:
(c² – ab)x² – 2(a² – bc)x + (b² – ac) – 0
Δ = B² – 4AC = 0 (since the roots are real and equal)
⇒ 4(a^2′ – bc)² – 4 (c² – ab)(b² – ac) = 0
⇒ 4(a⁴ – 2a²bc + b²c²) – 4(c²b² – ab³ – ac³ + a²bc) = 0
⇒ 4a⁴ + 4b²c² – 8a²bc – 4c²b² + 4ab³ + 4ac³ – 4a²bc = 0
⇒ 4a⁴+ 4ab³ + 4ac³ – 4a²bc – 8a²bc = 0
⇒ 4a [a³ + b³ + c³] = 0 or a = 0
⇒ a = 0 or [a³ + b³ + c³ – 3abc] = 0
⇒ a³ + b³ + c³ – 3abc = 0
⇒ a³ + b³ + c³ = 3abc or a = 0
Hence proved.

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