Question

find the following given in the picture​

Answer 1

We know that,

(x + y)² - (x - y)² = 4xy

Hence,

$\rm \dashrightarrow \Bigg ( a + \dfrac{1}{a} \Bigg )^2 -\Bigg ( a - \dfrac{1}{a}\Bigg )^2 = 4 \times a \times \dfrac{1}{a}$

From the Question,

• a + 1/a = 6

Hence,

$\rm \dashrightarrow ( 6)^2 -\Bigg ( a - \dfrac{1}{a}\Bigg )^2 = 4$

$\rm \dashrightarrow 36-\Bigg ( a - \dfrac{1}{a}\Bigg )^2 = 4$

$\rm \dashrightarrow -\Bigg ( a - \dfrac{1}{a}\Bigg )^2 = 4-36$

$\rm \dashrightarrow -\Bigg ( a - \dfrac{1}{a}\Bigg )^2 = -32$

$\rm \dashrightarrow \Bigg ( a - \dfrac{1}{a}\Bigg )^2 = 32$

$\rm \dashrightarrow a - \dfrac{1}{a} = \sqrt{32} = \sqrt{4\times 4\times 2}$

$\bf \dashrightarrow a - \dfrac{1}{a} = 4\sqrt{2} \qquad ..[1]$

We know that,

a² - b² = (a + b)(a - b)

Hence,

$\rm\dashrightarrow a^2-\dfrac{1}{a^2} = \Bigg ( a +\dfrac{1}{a} \Bigg )\Bigg ( a - \dfrac{1}{a}\Bigg )$

From the Question and [1]

• a + 1/a = 6

• a - 1/a = 4√2

Hence,

$\rm\dashrightarrow a^2-\dfrac{1}{a^2} =(6 ) ( 4\sqrt{2} )$

$\rm\dashrightarrow a^2-\dfrac{1}{a^2} = 6 \times 4 \sqrt{2}$

$\bf\dashrightarrow a^2-\dfrac{1}{a^2} = 24 \sqrt{2} \qquad ...[2]$