Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.14

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

Ex 3.14 Class 10 Samacheer Question 1.
Write each of the following expression in terms of α + β and αβ.

Solution:

10th Maths Exercise 3.14 Samacheer Kalvi Question 2.
The roots of the equation 2x² – 7x + 5 = 0 are α and β. Without solving the root find

Solution:
2x² – 7x + 5 = x² – \(\frac{7}{2} x+\frac{5}{2}\) = 0
α + β = \(\frac{7}{2}\)
αβ = \(\frac{5}{2}\)

Exercise 3.14 Class 10 Samacheer Question 3.
The roots of the equation x² + 6x – 4 = 0 are α, β. Find the quadratic equation whose roots are
(i) α² and β²
(ii) \(\frac{2}{\alpha} \text { and } \frac{2}{\beta}\)
(iii) α²β and β²α
Solution:
If the roots are given, the quadratic equation is x² – (sum of the roots) x + product the roots = 0.
For the given equation.
x² + 6x – 4 = 0
α + β = -6
αβ = -4
(i) α² + β² = (α + β)² – 2αβ
= (-6)² – 2(-4) = 36 + 8 = 44
α²β² = (αβ)² = (-4)² = 16
∴ The required equation is x² – 44x – 16 = 0.

(iii) α²β + β²α = αβ(α + β)
= -4(-6) = 24
α²β × β²α = α³β³ = (αβ)³ = (-4)³ = -64
∴ The required equation = x² – 24x – 64 – 0.

10th Maths Exercise 3.14 Question 4.
If α, β are the roots of 7x² + ax + 2 = 0 and if β – α = \(\frac{-13}{7}\) Find the values of a.
Solution:

10th Maths Exercise 3.14 Solutions Question 5.
If one root of the equation 2y² – 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

a² = 576
a = 24, -24

10th Maths Exercise 3.14 Solution Question 6.
If one root of the equation 3x² + kx + 81 = 0 (having real roots) is the square of the other then find k.
Solution:
3x² + kx + 81 = 0
Let the roots be α and α²