# Rational Exponent Equation.

• January 17th 2011, 12:43 PM
skoker
Rational Exponent Equation.
I am having some trouble with this one. I think exponentiation would work but i can't seem to get it right. the answer is x=36. could someone show an intermediate step so i can see which form, radical or fractional exponent to use. thanks.

$

{x}^{\frac {1} {2}}-6{x}^{\frac {-1}{2}}-5=0

$

$

\sqrt {x}-6\,{\frac {1} {\sqrt {x}}}}-5=0}

$
• January 17th 2011, 12:57 PM
e^(i*pi)
Multiply through by $x^{1/2}$

$x - 6 - 5\sqrt{x} = 0$ or, in standard form: $(\sqrt{x})^2 - 5\sqrt{x} - 6 =0$

This is a quadratic equation (and factors)

Edit: Note that the domain demands that $x > 0$ so discard any negative solution
• January 17th 2011, 02:45 PM
skoker
I tried to do a normal quadratic factor and get (x=4,x=9) which is not right. quadratic formula does not seem to give 36 also. I also tried to exponentiation but it just shifts the radical on x. how to factor this?
• January 17th 2011, 02:52 PM
e^(i*pi)
$(\sqrt{x}+1)(\sqrt{x}-6) = 0$ - using the quadratic formula would give you $\sqrt{x}$ so you'd need to square the value obtained to get x. Hence if you got 6 from the quadratic formula you'd have done it right

Perhaps it's easier if you let $u = \sqrt{x}$ which would give $u^2-5u-6=0$
• January 17th 2011, 02:56 PM
pickslides
From $(\sqrt{x})^2 - 5\sqrt{x} - 6 =0$

make $u= \sqrt{x}$

$u^2 - 5u - 6 =0$

$(u-6)(u+1) = 0$

$(\sqrt{x}-6)(\sqrt{x}+1) = 0$

Finish him off!
• January 17th 2011, 03:23 PM
skoker
ah I got it thanks. I got $(\sqrt{x}-2)(\sqrt{x}-3) = 0$ which looked right for a second. Is there a rule where you can't get 2 negative factors if the quadratic has negative terms, or something like that?
• January 17th 2011, 03:38 PM
e^(i*pi)
That's not right. The last term will be +6 instead of -6.

Me and Pickslides have given the correct factorisation in posts 4 and 5 respectively although I use trial and error to get it right when factoring
• January 17th 2011, 03:41 PM
skoker
yah. you are correct :) thanks for you help.
• January 17th 2011, 03:47 PM
evanator
You will have two values for $\sqrt{x}$.
You need to get rid of the negative one because for your purposes we say that the square root of a negative number is undefined. So there is one correct value for $\sqrt{x}$ here.

I think that's what you were asking about, but I'm not sure.