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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) 15x2 + 11x + 2 = 0
Here a = 15, b = 11, c = 2
∆ = b2 – 4ac
∆ = 112 – 4(15) × 2
= 121 – 120
∆ = 1 > 0
So the equation will have real and unequal roots

(ii) x2 – x – 1 = 0
Here a = 1, b = -1, c = -1
∆ = b2 – 4ac
= (-1)2 – 4(1)(-1)
= 1 + 4 = 5
∆ = 1 > 0
So the equation will have real and unequal roots.

(iii) $$\sqrt { 2 }$$ t2 – 3t + 3$$\sqrt { 2 }$$ = 0
Here a = $$\sqrt { 2 }$$ , b = -3, c = 3$$\sqrt { 2 }$$
∆ = b2 – 4ac
= (-3)2 – 4($$\sqrt { 2 }$$) (3$$\sqrt { 2 }$$)
= 9 – 24 = -15
∆ = -15 < 0
So the equation will have no real roots.

(iv) 9y2 – 6$$\sqrt { 2y }$$ + 2 = 0
Here a = 9, b = -6$$\sqrt { 2 }$$, c = 2
∆ = b2 – 4ac
= (-6$$\sqrt { 2 }$$)2 – 4(9) (2)
= 72 – 72
= 0
So the equation will have real and equal roots.

(v) 9a2b2x2 – 24abcdx + 16c2d2 = 0, a ≠ 0, b ≠ 0
Here a = 9a2b2; b = -24 abed, c = 16c2d2
∆ = b2 – 4ac
= (-24abcd)2 – 4(9a2b2) (16c2d2)
= 576a2b2c2d2 – 576a2b2c2d2
∆ = 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) kx2 + (6k + 2)x + 16 = 0
Solution:
(5k – 6)x2 + 2kx + 1 :
a = (5k – 6), b = 2, c = 1
Δ = b2 – 4ac
⇒ (2k)2 – 4 (5k – 6)(1)
⇒ 4k2 – 20k + 24 = 0 [∵ Since the roots are real and equal)
⇒ k2 – 5k + 6 = 0
⇒ (k – 3)(k – 2) = 0
k = 3, 2

(ii) kx2 + (6k + 2)x + 16 = 0
a = k, b = (6k + 2), c = 16
Δ = b2 – 4ac [∵ the roots are real and equal)
⇒ (6k + 2)2 – 4 × k × 16 = 0
⇒ 36k2 + 24k + 4 – 64k = 0
⇒ 36k2 – 40k + 4 =0
⇒ 36k2 – 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)x2 + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.
(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
∴ b2 – 4ac = 0
(b – c)2 – 4(a – b)(c – a) = 0
b2 + c2 – 2bc -4 (ac – a2 – bc + ab) = 0
b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
b2 + c2 + 2bc -4a (b + c) + 4a2 = 0
(b + c)2 – 4a (b + c) + 4a2 = 0
[(b+c) – 2a]2 = 0 [using a2 – 2ab + b2 = (a – b)2]
b + c – 2a = 0
b + c = 2a
b + c = a + a
c – a = a – b (t2 – t1 = t3 – t2)
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)x2 – 6(a + b)x – 9(a – b) = 0
Δ = b2 – 4ac
= (-6(a + b)2 – 4(a – b)(-9(a – b))
= 36(a + b)2 + 36(a – b)2
= 36 (a2 + 2ab + b2) + 36(a2 – 2ab + b2)
= 72a2 + 12b2
= 72(a2 + b2) > 0
∴ The roots are real and unequal.

Question 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
(c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0
Here a = c2 – ab ; b = – 2 (a2 – bc); c = b2 – ac
Since the roots are real and equal
∆ = b2 – 4ac
[-2 (a2 – bc)]2 – 4(c2 – ab) (b2 – ac) = 0
4(a2 – bc)2 – 4[c2 b2 – ac3 – ab3 + a2bc] = 0
Divided by 4 we get
(a2 – bc)2 – [c2 b2 – ac3 – ab3 + a2bc] = 0
a4 + b2 c2 – 2a2 bc – c2b2 + ac3 + ab3 – a2bc = 0
a4 + ab3 + ac3 – 3a2bc = 0
= a(a3 + b3 + c3) = 3a2bc
a3 + b3 + c3 = $$\frac{3 a^{2} b c}{a}$$
a3 + b3 + c3 = 3 abc
Hence it is proved