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pashok25 [27]
4 years ago
14

What are all of the real roots of the following polynomial? f(x) = x4 - 13x2 + 36

Mathematics
2 answers:
GREYUIT [131]4 years ago
5 0
ANSWER

A. -3, -2, 2, and 3

EXPLANATION

The given polynomial function is,

f(x) = {x}^{4} - 13 {x}^{2} + 36

To find the real roots, we equate the function to zero to obtain;

{x}^{4} - 13 {x}^{2} + 36 = 0

We can solve this equation as a quadratic equation in
{x}^{2}.

Thus we rewrite the equation as,

({x}^{2})^{2} - 13 {x}^{2} + 36 = 0

We split the middle term to get,

({x}^{2})^{2} - 9 {x}^{2} - 4 {x}^{2} + 36 = 0.

We factor to get,

{x}^{2} ( {x}^{2} - 9) - 4( {x}^{2} - 9) = 0

We factor to get,

( {x}^{2} - 9)( {x}^{2} - 4) = 0

Either

{x}^{2} - 9 = 0
or

{x}^{2} - 4 = 0

x = \pm \sqrt{9} \: or \: x = \pm \sqrt{4}
x = \pm3 \: or \: x = \pm2

x=-3,-2,2,3

The correct answer is A
Sergio [31]4 years ago
3 0
X^4 - 13x^2  + 36 = 0

(x^2 - 9)(x^2 - 4) = 0

x^2 - 9  = 0  gives x = 3, -3

x^2 - 4  = 0,  gives x = 2, -2.

Answer is A.
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