Solve Using substitution.

$x^{3/2}+8x^{1/2}+16x^{-1/2}=0$

$Let A= x^{1/2}$

$A^2+8A+\frac{16}{A}=0$

$A^3+8A^2+16=0$

$A^2(A+8)=-16$

I have no clue if i even did this correctly...Help me please.

Originally Posted by theloser
Solve Using substitution.

$x^{3/2}+8x^{1/2}+16x^{-1/2}=0$

$Let A= x^{1/2}$

$A^2+8A+\frac{16}{A}=0$
This should be

$A^3+8A+\frac{16}{A}=0$

Now multiply through by A.

So even if the the first factor is not ^2 of the second term, A substitution is still possible?

$A^4+8A^2+16=0$

$A^2(A^2+8)=-16$

$A=\pm\sqrt{-16}$
$Or$
$A=\pm\sqrt{-24}$

Originally Posted by theloser
So even if the the first factor is not ^2 of the second term, A substitution is still possible?

$A^4+8A^2+16=0$

$A^2(A^2+8)=-16$

$A=\pm\sqrt{-16}$
$Or$
$A=\pm\sqrt{-24}$
$A^4+8A^2+16=0$

Now $A^2 = B$ thus

$B^2+8B+16=0$

$(B+4)^2=0$ , back substitute twice.

$A=\pm\sqrt{-4}$

Originally Posted by theloser
$A=\pm\sqrt{-4}$
Yes, now you know that $\displaystyle x^{\frac{1}{2}} = \pm \sqrt{-4}$, what is $\displaystyle x$?

Not sure since there is already a $\pm$ on one side, how do I square both sides?

Originally Posted by theloser
Not sure since there is already a $\pm$ on one side, how do I square both sides?
What happens when you square a positive number? What happens when you square a negative number?

$x=\pm4$

$x=\pm-4$

Originally Posted by theloser
$x=\pm4$

$x=\pm-4$
Not even close, sorry.

$\displaystyle \left(\sqrt{-4}\right)^2 = \sqrt{-4} \times \sqrt{-4} = -4$.

$\displaystyle \left(-\sqrt{-4}\right)^2 = \left(-\sqrt{-4}\right)\left(-\sqrt{-4}\right) = \sqrt{-4}$.

Therefore $\displaystyle \left(\pm \sqrt{-4}\right)^2 = -4$.