Where is Square root of 1/5?

Algebra Level 3

$\sqrt{\frac{a^{100} - b^{100}}{a^{98}(a+1) - b^{98}(b+1)}}$ Let $a,b$ be the roots of $x^2 - x -1 = 0$ , then find the value of the above expression.

The answer is 1.

1 solution

Rishabh Jain
Mar 5, 2016

$x^2-x-1=0\implies x^2=x+1$ and since $a,b$ are its roots, $\therefore a^2=a+1\text{ and } b^2=b+1$ Hence, $\sqrt{\frac{a^{100} - b^{100}}{a^{98}(a+1) - b^{98}(b+1)}} =\sqrt{\frac{a^{100} - b^{100}}{a^{98}(a^2) - b^{98}(b^2)}}~~$ $\Large ~~~~~~~~~~~=\sqrt 1=\huge \boxed 1$

Rishab cool returns! Though how the hell is this question lvl 5?

Shreyash Rai - 5 years, 3 months ago

Is it not level 3?

Sean O'Neill - 5 years, 3 months ago

When I solved it, it was lvl 5

Shreyash Rai - 5 years, 3 months ago

I also couldn't understand why this question is level 5 ??

Rishabh Jain - 5 years, 3 months ago

Yes! You are right SHREYAS.

Atanu Ghosh - 5 years, 3 months ago

Where is Square root of 1/5?

The answer is 1.

1 solution

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