What Weird Roots!

Algebra Level 3

If the roots of $x^{2} + qx + p = 0$ are $\sqrt{6 + 2\sqrt{5}}$ and $\sqrt{6 - 2\sqrt{5}},$ what is $p - q?$

2

4 - 2\sqrt{5}

6

4 + 2\sqrt{5}

-6

2 solutions

Sakanksha Deo
Mar 10, 2015

$\sqrt{ 6 + 2\sqrt{5} } = \sqrt{ ( \sqrt{5} )^{2} + 1^{2} +2\times1\times\sqrt{5} } = \sqrt{ ( \sqrt{5} + 1 )^{2} } = (\sqrt{5} + 1)$

$\sqrt{ 6 - 2\sqrt{5} } = \sqrt{ ( \sqrt{5} )^{2} + 1^{2} -2\times1\times\sqrt{5} } = \sqrt{ ( \sqrt{5} - 1 )^{2} } = (\sqrt{5} - 1)$

Sum of roots = $-q = \sqrt{5} + 1 + \sqrt{5} - 1 = 2\sqrt{5}$

Product of roots = $p = ( \sqrt{5} + 1 ) ( \sqrt{5} - 1 ) = 5 - 1 = 4$

Therefore,

$p - q = 4 + 2\sqrt{5}.$

Yes, one should always consider that the radicals could be simplified.

Damn it!! I took the sum to be =q.. Sob Sob

Mehul Arora - 6 years, 2 months ago

Don't worry....such things can happen. What's important, is to learn from your mistakes... Try other questions of my set

Sakanksha Deo - 6 years, 2 months ago

Yup! Thanks for the motivation.

Mehul Arora - 6 years, 2 months ago

in the second equatio why cant it be 1-√5

Ranjith Vuppala - 4 years, 10 months ago

$1- \sqrt 5$ is negative, and the square root of a negative number is not real. Thus it is taken to be $\sqrt 5 -1$

Mehul Arora - 4 years, 10 months ago

Jason Zou
Jun 30, 2015

$p$ is the product of the roots. $p=\sqrt{(6+2\sqrt{5})(6-2\sqrt{5})}=\sqrt{16}=4$

$-q$ is the sum of the roots. We have

$q^2=6+2\sqrt{5}+6-2\sqrt{5}+2\sqrt{(6+2\sqrt{5})(6-2\sqrt{5})}=12+2p=20$

Becuase both roots are positive, the sum of the roots must also be positive, so $-q=2\sqrt{5}$

So now we have $p-q=\boxed{4+2\sqrt5}$