02 Apr 2022

Samacheer kalvi 10th Maths – Algebra Ex 3.13

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

Samacheer Kalvi 10th Standard Maths Book Back Questions with Answers PDF uploaded and the same given below. Class-tenth candidates and those preparing for TNPSC exams can check the Maths Book Back Answers PDF below. Samacheer Kalvi Class 10th Std Maths Book Back Answers Chapter 3 Exercise 3.13 Solutions are available below. Check the complete Samacheer Kalvi 10th Maths – Algebra Ex 3.13 Book Back Answers below:

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

Samacheer Kalvi 10th Maths Book Subject One Mark, Two Mark, Five Mark Guide questions and answers are below. Check Maths Book Back Questions with Answers. Take the printout and use it for exam purposes.

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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.
10th maths unit - 3 book back answer

∴ 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
10th maths unit - 3 book back answer

∆ = 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(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.

Other Important Links for 10th Maths Book Back Answers Solutions:

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