Question 109416
"Solve:x^4-9x^2+18=0"


Let w=x^2 to get 


{{{w^2-9w+18=0}}}



Let's use the quadratic formula to solve for w:



Starting with the general quadratic


{{{aw^2+bw+c=0}}}


the general solution using the quadratic equation is:


{{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{w^2-9*w+18=0}}} ( notice {{{a=1}}}, {{{b=-9}}}, and {{{c=18}}})





{{{w = (--9 +- sqrt( (-9)^2-4*1*18 ))/(2*1)}}} Plug in a=1, b=-9, and c=18




{{{w = (9 +- sqrt( (-9)^2-4*1*18 ))/(2*1)}}} Negate -9 to get 9




{{{w = (9 +- sqrt( 81-4*1*18 ))/(2*1)}}} Square -9 to get 81  (note: remember when you square -9, you must square the negative as well. This is because {{{(-9)^2=-9*-9=81}}}.)




{{{w = (9 +- sqrt( 81+-72 ))/(2*1)}}} Multiply {{{-4*18*1}}} to get {{{-72}}}




{{{w = (9 +- sqrt( 9 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{w = (9 +- 3)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{w = (9 +- 3)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{w = (9 + 3)/2}}} or {{{w = (9 - 3)/2}}}


Lets look at the first part:


{{{x=(9 + 3)/2}}}


{{{w=12/2}}} Add the terms in the numerator

{{{w=6}}} Divide


So one answer is

{{{w=6}}}




Now lets look at the second part:


{{{x=(9 - 3)/2}}}


{{{w=6/2}}} Subtract the terms in the numerator

{{{w=3}}} Divide


So another answer is

{{{w=3}}}


So our solutions are:

{{{w=6}}} or {{{w=3}}}



Remember, we let {{{w=x^2}}} so 


{{{x^2=6}}} or {{{x^2=3}}}


So our answers in terms of x are 



*[Tex \LARGE x=\pm\sqrt{6}] or  *[Tex \LARGE x=\pm\sqrt{3}] 



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"Find k so that 4x^2-kx+1=0 has one rational solution. "



Remember, if the discriminant {{{b^2-4ac}}} is equal to zero, then the quadratic will have one real solution



{{{k^2-4(4)(1)=0}}} Set the discriminant equal to zero



{{{k^2-16=0}}} Multiply



{{{k^2=16}}} Add 16 to both sides



*[Tex \LARGE k=\pm\sqrt{16}] Take the square root of both sides



*[Tex \LARGE k=\pm 4] Simplify


So the values of k are {{{k=-4}}} or {{{k=4}}}