Question

if a = 1/a-5, find a²- 1/a² please tell correctly​

Answer 1

Topic :-

Algebraic Identities

Given :-

$\sf{a=\dfrac{1}{a}-5}$

To Find :-

$\sf {a^2-\dfrac{1}{a^2}}$

Solution :-

We need to find,

$\sf {a^2-\dfrac{1}{a^2}}$

which can be written as,

$\sf{\left(a+\dfrac{1}{a} \right)\left(a-\dfrac{1}{a} \right)}$

$\sf{(\because x^2-y^2=(x+y)(x-y))}$

We are provided with,

$\sf{a=\dfrac{1}{a}-5}$

which can be written as,

$\sf{a-\dfrac{1}{a}=-5}$

So, we need to find,

$\sf{\left(a+\dfrac{1}{a} \right)}$

to answer the required question.

So,

$\sf{a-\dfrac{1}{a}=-5}$

Squaring both sides,

$\sf{\left(a-\dfrac{1}{a}\right)^2=(-5)^2}$

$\sf{a^2-2(a)\left(\dfrac{1}{a}\right)+\left(\dfrac{1}{a}\right)^2=(-5)^2}$

$\sf {(\because (x-y)^2=x^2-2xy+y^2)}$

$\sf{a^2-2+\dfrac{1}{a^2}=25}$

$\sf{a^2+\dfrac{1}{a^2}=25+2}$

$\sf{a^2+\dfrac{1}{a^2}=27}$

Now,

$\sf{\left( a+\dfrac{1}{a}\right)^2=a^2+2(a)\left( \dfrac{1}{a}\right)+\left( \dfrac{1}{a}\right)^2}$

$\sf {(\because (x+y)^2=x^2+2xy+y^2)}$

$\sf {\left( a+\dfrac{1}{a}\right)^2=a^2+2+\dfrac{1}{a^2}}$

$\sf {\left( a+\dfrac{1}{a}\right)^2=\left(a^2+\dfrac{1}{a^2}\right)+2}$

$\sf {\left( a+\dfrac{1}{a}\right)^2=27+2}$

$\sf {\left(\because a^2+\dfrac{1}{a^2}=27\right)}$

$\sf {\left( a+\dfrac{1}{a}\right)^2=29}$

Taking Square root both sides,

$\sf {\sqrt{\left( a+\dfrac{1}{a}\right)^2}=\sqrt{29}}$

$\sf {a+\dfrac{1}{a}=\pm\sqrt{29}}$

Now, calculating the required value,

$\sf{\left(a+\dfrac{1}{a} \right)\left(a-\dfrac{1}{a} \right)}$

$\sf {(\pm\sqrt{29})(-5)}$

$\sf {\left(\because a+\dfrac{1}{a}=\pm\sqrt{29}\right)}$

$\sf {\left(\because a-\dfrac{1}{a}=-5\right)}$

$\sf {\mp\:5\sqrt{29}}$

Answer :-

$\underline{\boxed{\sf {a^2-\dfrac{1}{a^2}=\mp\:5\sqrt{29}}}}$