Thread: Not sure what this comes under.

1. Not sure what this comes under.

Had this question in a test and got it wrong.

find the solutions to $\displaystyle 2\sqrt{x^5}-72\sqrt{x}=0$

answer is x = 0 , x=6 , or x = -6; not sure what i do with this equation , im on a study break at moment so im trying to get on top of these things before i get back. You help is appreciated

2. Originally Posted by el123
Had this question in a test and got it wrong.

find the solutions to $\displaystyle 2\sqrt{x^5}-72\sqrt{x}=0$

answer is x = 0 , x=6 , or x = -6; not sure what i do with this equation , im on a study break at moment so im trying to get on top of these things before i get back. You help is appreciated
I don't know where you got "x= 0, x= 6, or x=-6" but 6 and -6 are obviously NOT solutions. If x= 6, for example, then $\displaystyle x^5= 6^5= 7776$ so $\displaystyle \sqrt{x^5}= \sqrt{7776}= 88.816$ while [tex]\sqrt{6}= 2.44945 so $\displaystyle 2\sqrt{x^5}- 72\sqrt{x}$$\displaystyle = 2(88.816)- 72(2.44945)= 1.2716$, not 0. Did you mean "x= 0, $\displaystyle x^2= 6$, or $\displaystyle x^2= -6$"?

$\displaystyle \sqrt{x^5}= \left(\sqrt{x}\right)^5$ so let $\displaystyle u= \sqrt{x}$ and the equation becomes $\displaystyle 2u^5- 72u= 0$. That can be factored as $\displaystyle 2u(u^4- 36)= 2u(u^2- 6)(u^2+ 6)= 0$. Therefore, we have u= 0 or $\displaystyle u^2- 6= 0$ or $\displaystyle u^2+ 6= 0$. The 5 solutions to this fifth degree equation are x= 0, $\displaystyle x= \sqrt{6}$, $\displaystyle x= -\sqrt{6}$, $\displaystyle x= i\sqrt{6}$, and $\displaystyle x= -i\sqrt{6}$.

3. Were not using imaginary numbers in this course , and the answer is the one i gave because we have been given the answers for it.

4. Actually, x=0,-6,6 are solutions.

How?

$\displaystyle 2\sqrt{x^5}=72\sqrt{x}$
$\displaystyle \sqrt{x^5}=36\sqrt{x}$
$\displaystyle x^5=1296x$
$\displaystyle x^4=1296$
$\displaystyle x=6,-6$
And $\displaystyle x=0$ is an obvious solution.

Put into the equation
$\displaystyle 2\sqrt{x^5}=176.363...$
$\displaystyle 72\sqrt{6}=176.363...$
Hence:$\displaystyle 2\sqrt{6^5}-72\sqrt{6}=0$

Actually, your values of $\displaystyle x=\sqrt{6}$ are wrong
$\displaystyle 2\sqrt{\sqrt{6}}=3.130...$
$\displaystyle 72\sqrt{\sqrt{6}}=112.6860...$
And so on with your other values

HallsofIvy, here's your mistake
$\displaystyle 2u(u^2-6)(u^2+6)=0$
$\displaystyle u^2-6=0$
$\displaystyle u=\sqrt{6}$
$\displaystyle u=\sqrt{x}$
Hence
$\displaystyle \sqrt{x}=\sqrt{6}$
$\displaystyle x=6$
And so on

5. Originally Posted by I-Think
Actually, x=0,-6,6 are solutions.
x = -6 is not a solution ... for real solutions, the domain of the equation is $\displaystyle x \ge 0$