Mean of the Solutions

Algebra Level 4

Find the mean of all of the real solutions of: $x^2+x\sqrt{x}=12x$

-0.5 12.5 9 4.5

1 solution

Michael Ng
May 23, 2014

First we move everything to one side and factorise: $x(x + \sqrt{x} - 12) = 0$ $x(\sqrt{x}+4)(\sqrt{x}-3)=0$

Hence $\sqrt{x}$ = -4 or 3, or x = 0. But $\sqrt{x}$ cannot be -4 by definition, hence x can only be 0 or 9. Therefore the answer is 4.5.

I personally think that a choice should have been 25/3...even I was confused for a second when I didn't see one!

Nathan Ramesh - 6 years, 11 months ago

Why can't $\sqrt{x}$ be $-4$ ?

Nanayaranaraknas Vahdam - 7 years ago

Because only $\pm\sqrt{x}$ can be -4 or 4, not $\sqrt{4}$ . There's no option for the mean of 0, 16 and 9, so that avoids confusion.

Michael Ng - 7 years ago

i took $\sqrt{x}=a$ so this will give you $x=9 or 16$

g j - 6 years, 11 months ago

@G J – X cannot be 16 because it is an extraneous root.

Kushagra Sahni - 6 years, 11 months ago

Because the question asks for "real" solutions.

Nishant Sharma - 7 years ago

you got me there :3

Md Lokman Hosen - 4 years, 8 months ago