You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.14

**10th Maths Exercise 3.14 Solutions Question 1.**

Write each of the following expression in terms of α + β and αβ.

Solution:

**Ex 3.14 Class 10 Samacheer Question 2.**

The roots of the equation 2x^{2} – 7x + 5 = 0 are α and β. Without solving the root find

Solution:

2x^{2} – 7x + 5 = x^{2} – \(\frac{7}{2} x+\frac{5}{2}\) = 0

α + β = \(\frac{7}{2}\)

αβ = \(\frac{5}{2}\)

**10th Maths Exercise 3.14 Question 3.**

The roots of the equation x^{2} + 6x – 4 = 0 are α, β. Find the quadratic equation whose roots are

(i) α^{2} and β^{2}

(ii) \(\frac{2}{\alpha} \text { and } \frac{2}{\beta}\)

(iii) α^{2}β and β^{2}α

Solution:

If the roots are given, the quadratic equation is x^{2} – (sum of the roots) x + product the roots = 0.

For the given equation.

x^{2} + 6x – 4 = 0

α + β = -6

αβ = -4

(i) α^{2} + β^{2} = (α + β)^{2} – 2αβ

= (-6)^{2} – 2(-4) = 36 + 8 = 44

α^{2}β^{2} = (αβ)^{2} = (-4)^{2} = 16

∴ The required equation is x^{2} – 44x – 16 = 0.

(iii) α^{2}β + β^{2}α = αβ(α + β)

= -4(-6) = 24

α^{2}β × β^{2}α = α^{3}β^{3} = (αβ)^{3} = (-4)^{3} = -64

∴ The required equation = x^{2} – 24x – 64 – 0.

**10th Maths Exercise 3.14 Samacheer Kalvi Question 4.**

If α, β are the roots of 7x^{2} + ax + 2 = 0 and if β – α = \(\frac{-13}{7}\) Find the values of a.

Solution:

**10th Maths Exercise 3.14 Solution Question 5.**

If one root of the equation 2y^{2} – ay + 64 = 0 is twice the other then find the values of a.

Solution:

Let one of the root α = 2β

α + β = 2β + β = 3β

Given

a^{2} = 576

a = 24, -24

**Exercise 3.14 Class 10 Samacheer Question 6.**

If one root of the equation 3x^{2} + kx + 81 = 0 (having real roots) is the square of the other then find k.

Solution:

3x^{2} + kx + 81 = 0

Let the roots be α and α^{2}