Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Students can download Maths Chapter 3 Algebra Ex 3.13 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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
Answer:
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
Answer:
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.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

(iii) \(\sqrt { 2 }\) t2 – 3t + 3\(\sqrt { 2 }\) = 0
Answer:
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
Answer:
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
Answer:
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.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Question 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.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13 1
∴ 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
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13 2
∆ = 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 = \(\frac { 1 }{ 9 } \)
The value of k = 1 or \(\frac { 1 }{ 9 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

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.
Answer:
(a – b) x2 + (b – c) x + (c – a) = 0
Here a = (a – b);b = b – c ; c = c – a
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13 3
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.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Question 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.
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.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

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
Answer:
(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

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