# 10th Maths Book Back Algebra Ex 3.13

## Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 3 – Algebra Ex 3.13 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 3 Algebra consists of 19 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Algebra Ex 3.13 Answers in English.

### 10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Unit 3 Chapter 13 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,

### 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
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
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, 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.
(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(a2′ – 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.