Question 122517
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Let {{{x=b^2}}}, then {{{x^2=b^4}}}.

Substitute:
{{{b^4+13b^2+36=0}}}
{{{x^2+13x+36=0}}}

The quadratic factors:
{{{(x+4)(x+9)=0}}}

Therefore
{{{x=-4}}} or {{{x=-9}}}

But {{{x=b^2}}} so

{{{b^2=-4}}} or {{{b^2=-9}}}

Therefore

{{{b=2i}}}
{{{b=-2i}}}
{{{b=3i}}}
{{{b=-3i}}}

Giving us the 4 expected roots for the quartic (degree 4) equation.

Since the roots are all complex, there are no real zeros for this equation,
therefore the graph will not intercept the x-axis anywhere.


{{{graph(600,600,-5,5,-5,80,x^4+13x^2+36)}}}

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