10th Maths - Book Back Answers - Chapter 3 Exercise 3.13 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 3 Exercise 3.13 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 3 Exercise 3.13: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 Exercise 3.13 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 3 Exercise 3.13 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 3 Algebra Ex 3.13

1. Determine the nature of the roots for the following quadratic equations
(i) 15x² + 11x + 2 = 0
Answer Key:
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 Key:
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
Answer Key:
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$\sqrt{2\mathrm{y}}$+ 2 = 0
Answer Key:
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 Key:
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.

 

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

 

(i) (5k – 6)² + 2kx + 1 = 0
Answer Key:
(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

Answer Key:

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 = 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.
Answer Key:
(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.

 

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.
Answer Key:
(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.

 

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 Key:
(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 = $\frac{3a^{2}bc}{a}$
a³ + b³ + c³ = 3 abc
Hence it is proved