Question 84579
The method you are asked to use for this problem is the method of substitution.  You should begin by noticing that in this equation, if you let {{{u=x^2}}}, then {{{u^2 = (x^2)^2= x^4}}}


Now, make these substitutions and you get:
{{{x^4 - 14x^2 - 32 = 0}}}
{{{u^2 - 14u - 32 = 0 }}}


This factors into
{{{(u-16)(u+2) = 0}}}
{{{u=16}}} or {{{u= -2}}}


Don't stop here!  You must still solve for the original variable which is x.  Substitute these values of u back into the formula {{{u=x^2}}}

{{{u=16}}} or {{{u= -2}}}
{{{x^2=16}}} or {{{x^2= -2}}}


Now, are you supposed to give REAL solutions, or are you supposed to also include imaginary (complex) solutions?  IF COMPLEX solutions are allowed, then
{{{x^2=16}}} or {{{x^2= -2}}}
{{{x= 0+-sqrt(16)}}} or {{{x= 0+-sqrt(-2) }}}
{{{x= 4}}}, {{{x=-4}}} or {{{x= 0+-i*sqrt(2) }}}


If you are finding ONLY REAL solutions, then the answer is only x=4 or x=-4.


Note:  You don't really need the 0 in the answers above.  It's just that in the algebra.com format, you can't write this + or - notation without using the 0!  Sorry about that.


R^2 at SCC