Should you factorize?

Algebra Level 3

You are given that

a − 1 a = a a 2 + 1 , a - \dfrac{1}{a} = \dfrac{a}{a^2+1} ,

find the value of

1 a 2 − a 2 . \dfrac{1}{a^2}-a^2 .


The answer is -1.

Viki Zeta by Viki Zeta , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Viki Zeta
Sep 6, 2016

a − 1 a = a a 2 + 1 = 1 a 2 + 1 a = 1 a + 1 a 1 a 2 − a 2 = − ( a 2 − 1 a 2 ) = − ( ( a + 1 a ) × ( a − 1 a ) ) = − ( ( a + 1 a ) × ( 1 a + 1 a ) ) = − 1 a - \dfrac{1}{a} = \dfrac{a}{a^2+1} = \dfrac{1}{\dfrac{a^2+1}{a}} = \dfrac{1}{a + \dfrac{1}{a}} \\ \dfrac{1}{a^2} - a^2 = -(a^2 - \dfrac{1}{a^2}) \\ = -((a+\dfrac{1}{a}) \times (a - \dfrac{1}{a})) \\ = - ( (a+\dfrac{1}{a}) \times (\dfrac{1}{a+\dfrac{1}{a}}) ) \\ = -1

Therefore the answer is -1 \fbox{ -1 }

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I'm getting 1,can u Check ur solution.

naitik sanghavi - 4 years, 9 months ago

Log in to reply

Sorry, this edit feature is not working properly. I once edited it and it didn't change. Open a new report to make those who answered 1 before this edit as correct.

Viki Zeta - 4 years, 9 months ago
Munem Shahriar
Jul 27, 2018

a − 1 a = a a 2 + 1 ⇒ a 2 − 1 a = a a 2 + 1 ⇒ ( a 2 − 1 ) ( a 2 + 1 ) = a 2 ⇒ a 4 − a 2 + a 2 − 1 = a 2 ⇒ a 4 − a 2 = 1 ⇒ a 2 ( a 2 − 1 ) = 1 ⇒ a 2 − 1 = 1 a 2 ⟹ 1 a 2 − a 2 = − 1 \begin{aligned} a - \dfrac 1a & = \dfrac a{a^2 +1} \\ \Rightarrow \dfrac{a^2 -1}a & = \dfrac a{a^2 + 1} \\ \Rightarrow (a^2-1)(a^2+1) & = a^2 \\ \Rightarrow a^4 - a^2 + a^2 - 1 & = a^2 \\ \Rightarrow a^4 - a^2 & = 1 \\ \Rightarrow a^2(a^2 -1) & = 1 \\ \Rightarrow a^2 -1 & = \dfrac 1{a^2} \\ \implies \dfrac 1{a^2} - a^2 & = \boxed{-1} \\ \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...