Question 1013018
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Find four roots of the following equation:

2x^4 +5x^2 = 207
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Introduce new variable for your convenience, y = {{{x^2}}}.

Then your equation takes the form

{{{2y^2 + 5y - 207}}} = {{{0}}}.

Apply the quadratic formula. You will get

{{{y[1,2]}}} = {{{(-5 +- sqrt(25 +4*2*207))/4)}}} = {{{(-5 +- sqrt(1681))/4}}} = {{{(-5 +- 41)/4}}}.

So, {{{y[1]}}} = 9, {{{y[2]}}} = {{{-23/2}}}.

Now you need to solve this two quadratic equations:

1)  {{{x^2}}} = 9. It has two solutions:  x = {{{3}}}  and  x = {{{-3}}}.

2)  {{{x^2}}} = {{{-23/2}}}. It has two solutions:  x = {{{i*sqrt(23/2)}}}  and  x = {{{-i*sqrt(23/2)}}}.

Thus you got four solutions of your original equation.
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