Question

x²+(a+1/a)x+1=0 by completing square method​

Answer 1

Answer:

- a, - 1/a are the roots of the equation.

Step-by-step explanation:

Given Equation :

$\sf \Rightarrow x^{2} + \bigg(a + \dfrac{1}{a} \bigg) + 1 = 0$

Transposing 1 to RHS

$\sf \Rightarrow x^{2} + \bigg(a + \dfrac{1}{a} \bigg) = - 1$

It can be written as

$\sf \Rightarrow x^{2} + 2\bigg( \dfrac{a}{2} + \dfrac{1}{2a} \bigg) = - 1$

Adding ( a/2 + 1/2a )² on both sides

$\sf \Rightarrow x^{2} + 2\bigg( \dfrac{a}{2} + \dfrac{1}{2a} \bigg) + \bigg( \dfrac{a}{2} + \dfrac{1}{2a} \bigg)^{2} = - 1 + \bigg( \dfrac{a}{2} + \dfrac{1}{2a} \bigg)^{2}$

Since ( p + q )² = p² + q² + 2pq

$\sf \Rightarrow \bigg(x + \dfrac{a}{2} + \dfrac{1}{2a} \bigg)^{2} = - 1 + \bigg( \dfrac{a}{2} \bigg) ^{2} + \bigg(\dfrac{1}{2a} \bigg)^{2} + \dfrac{1}{2}$

$\sf \Rightarrow \bigg(x + \dfrac{a}{2} + \dfrac{1}{2a} \bigg)^{2} = \bigg( \dfrac{a}{2} \bigg) ^{2} + \bigg(\dfrac{1}{2a} \bigg)^{2} - \dfrac{1}{2}$

To make a perfect square on RHS write 1/2 as 2( a/2 )( 1/2a )

$\sf \Rightarrow \bigg(x + \dfrac{a}{2} + \dfrac{1}{2a} \bigg)^{2} = \bigg( \dfrac{a}{2} \bigg) ^{2} + \bigg(\dfrac{1}{2a} \bigg)^{2} - 2 \bigg(\dfrac{a}{2} \bigg) \bigg( \dfrac{1}{2a} \bigg)$

Using identity p² + q² - 2pq = ( p - q )²

$\sf \Rightarrow \bigg(x + \dfrac{a}{2} + \dfrac{1}{2a} \bigg)^{2} = \bigg( \dfrac{a}{2} - \dfrac{1}{2a} \bigg) ^{2}$

Taking square root on both sides

$\sf \Rightarrow x + \dfrac{a}{2} + \dfrac{1}{2a} = \pm \bigg( \dfrac{a}{2} - \dfrac{1}{2a} \bigg)$

$\sf \Rightarrow x + \dfrac{a}{2} + \dfrac{1}{2a} = \dfrac{a}{2} - \dfrac{1}{2a} \qquad OR \qquad x + \dfrac{a}{2} + \dfrac{1}{2a} = - \bigg( \dfrac{a}{2} - \dfrac{1}{2a} \bigg)$

$\sf \Rightarrow x + \dfrac{1}{2a} = - \dfrac{1}{2a} \qquad OR \qquad x + \dfrac{a}{2} = - \dfrac{a}{2}$

$\sf \Rightarrow x = - \dfrac{1}{2a} - \dfrac{1}{2a} \qquad OR \qquad x = - \dfrac{a}{2} - \dfrac{a}{2}$

$\sf \Rightarrow x = - \dfrac{1}{a} \qquad OR \qquad x = -a$

Hence the roots of the equation are - a and - 1/a.