10th Maths - Book Back Answers - Chapter 3 Exercise 3.14 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 3 Exercise 3.14 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 3 Exercise 3.14: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 Exercise 3.14 – 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 3.14 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 3.14

1. Write each of the following expression in terms of α + β and αβ
(i)$\frac{\alpha }{3\beta }+\frac{\beta }{3\alpha }$

Answer Key: 

(ii) $\frac{1}{{\alpha }^2\beta }$+$\frac{1}{{\beta }^2\alpha }$
Answer Key: 

(iii) (3α – 1) (3β – 1)
Answer Key:
(3α – 1) (3β – 1) = 9αc – 3α – 3β + 1
= 9αβ – 3(α + β) + 1

 

(iv)$\frac{\alpha +3}{\beta }+\frac{\beta +3}{\alpha }$
Answer Key: 

 

2. The roots of the equation 2x² – 7x + 5 = 0 are a and p. Find the value of [without solving the equation]
 (i)$\frac{1}{{\alpha }^{}}$+$\frac{1}{{\beta }^{}}$
Answer Key: 
 α and α are the roots of the equation 2x² – 7x + 5 = 0
α + β = $\frac{7}{2}$ ; αβ = $\frac{5}{2}$
(i) $\frac{1}{{\alpha }^{}}$+$\frac{1}{{\beta }^{}}$=$\frac{\beta +\alpha }{\alpha \beta }$
 =$\frac{7}{2}+\frac{5}{2}=\frac{7}{2}\times \frac{2}{5}=\frac{7}{5}$

 

(ii) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$
Answer Key: 

= ($\frac{7}{2}$)² – 2 × $\frac{5}{2}$ ÷ $\frac{5}{2}$
= $\frac{49}{4}$ – 5 ÷ $\frac{5}{2}$
 =$\frac{49-20}{4}$÷ $\frac{5}{2}$
= $\frac{29}{4}$× $\frac{2}{5}$ =$\frac{29}{10}$

 

(iii) $\frac{\alpha +2}{\beta +2}+\frac{\beta +2}{\alpha +2}$
Answer Key: 

3. The roots of the equation x² + 6x – 4 = 0 are a, p. Find the quadratic equation whose roots are
(i) α² and β²
Answer Key: 
 α and β are the roots of x² + 6x – 4 = 0
α + β = -6; αβ = -4 

(i) Sum of the roots = α² + β² 
= (α + β)² – 2αβ
= 36 – 2 – (4) = 36 + 8
= 44
Product of the roots = α² + β²
= (αβ)²
= (-4)²
= 16
The Quadratic equation is
x² – (sum of the roots) x + Product of the roots = 0

x² – (44)x + 16 = 0
x² – 44x + 16 = 0

 

(ii) $\frac{2}{\alpha }$and$\frac{2}{\beta }$
Answer Key: 
 Sum of the roots =$\frac{2}{\alpha }$+$\frac{2}{\beta }$
 =$\frac{2\alpha +2\beta }{\alpha \beta }=\frac{2(\alpha +\beta )}{\alpha \beta }$
= $\frac{2\alpha +2\beta }{\alpha \beta }=\frac{2(\alpha +\beta )}{\alpha \beta }$=3
Product of the roots = $\frac{2}{\alpha }\times \frac{2}{\beta }=\frac{4}{\alpha \beta }$
= $\frac{4}{-4}=-1$
The Quadratic equation is
x² – (sum of the roots) x + Product of the roots = 0
x² – 3x – 1 = 0

(iii) α²β and β²α
Answer Key: 
 Sum of the roots = α²β + β²α
= αβ (α + β)
= -4 (-6) = 24
Product of the roots = α²β × β²α
= α²β³ = (αβ)³
= (-4)³ = -64
The Quadratic equation is
x² – (Sum of the roots) x + Product of the roots = 0
x² – 24x – 64 = 0

 

4. If α, β are the roots of 7x² + ax + 2 = 0 and if β – α = $\frac{13}{7}$ Find the values of a.
Answer Key: 

 α and β are the roots of 7x² + ax + 2 = 0
α + β = $\frac{-a}{7}$; αβ =$\frac{2}{7}$
Given β – α = – $\frac{13}{7}$⇒ α – β =$\frac{13}{7}$
Squaring on both sides
(α – β)² = ($\frac{13}{7}$)²
α² + β² = 2αβ = $\frac{169}{49}$
($\frac{-a}{7}$)² -4($\frac{2}{7}$) = $\frac{169}{49}$ ⇒ $\frac{a^2}{49}-\frac{8}{7}$=$\frac{169}{49}$
$\frac{a^2}{49}$ =$\frac{225}{49}$ ⇒ a2 =$\frac{\mathrm{225x49}}{49}$
a² = 225 ⇒ a = ± $\sqrt{225}$= ± 15
The value of a = 15 or – 15

 

5. If one root of the equation 2y², – ay + 64 = 0 is twice the other then find the values of a.
Answer Key: 
Let the roots be α and 2α
Here a = 2, b = – a, c = 64
Sum of the roots = – $\frac{b}{a}$
α + 2α =$\frac{a}{2}$
3α =$\frac{a}{2}$
a = 6α …….(1)
Product of the roots =$\frac{c}{a}$
α × 2α = $\frac{64}{2}$ = 2α² = 32
α² = $\frac{32}{2}$ = 16
α = $\sqrt{16}$= ± 4
Substitute the value of a in (1)
When α = 4
a = 6(4)
a = 24
The Value of a is 24 or -24
When α = -4
a = 6(-4)
a = -24

 

6. If one root of the equation 3x² + kx + 81 = 0 (having real roots) is the square of the other then find k.
Answer Key: 
 Let α and α² be the root of the equation 3x² + kx + 81
Here a = 3, b = k, c = 81
Sum of the roots = –$\frac{b}{a}$ = –$\frac{k}{\mathit{3}}$
α + α² = –$\frac{k}{\mathit{3}}$
3α + 3α² = -k ……..(1)
Product of the roots = $\frac{c}{a}=\frac{81}{3}$= 27
α × α² = 27
α³ = 27 ⇒ α³ = 3³
α = 3
Substitute the value of α = 3 in (1)
3(3) + 3(3)² = -k
9 + 27 = -k ⇒ 36 = – k
∴ k = -36
The value of k = -36