Question 165809
{{{x^4-81}}} is the same thing as saying {{{(x^2)^2-9^2}}}  You may be familiar with the special rule for factoring "the difference of 2 perfect squares"
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The difference of two perfect squares is as follows:
{{{a^2-b^2=(a-b)(a+b)}}}
this can be proven true by foiling (a-b)(a+b)={{{a^2+ab-ab-b^2}}}={{{a^2-b^2}}}
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So...in this equation
{{{(x^2)^2-9^2=0}}}  
{{{a=x^2}}} and {{{b=9}}}
so (x^2-9)(x^2+9)=0
the Zero-Product Property says that when 2 numbers are multiplied together, if either number is zero, then the product will be zero.
so
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the equation will be true when (x^2-9)=0 or when (x^2+9)=0
{{{x^2-9=0}}}
add 9 to each side
{{{x^2=9}}}
take the square root of each side
{{{x=+-3}}}
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{{{x^2+9=0}}}
Subtract 9 from each side
{{{x^2=-9}}}
take the square root of each side
{{{x=SQRT(-9)}}} since we are unable to take the square root of a negative number, this is not a real solution.
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so
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The final answer is {{{x=+-3}}} but the question is asking for all of the above work.
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