Math

QuestionFind the real zeros of P(x)=x44x296P(x)=x^{4}-4x^{2}-96. What are the xx-intercepts?

Studdy Solution

STEP 1

Assumptions1. We are asked to find the real zeros of the function, which are the values of xx that make the function (x)(x) equal to zero. . The real zeros of the function are also the x-intercepts of the graph of the function.

STEP 2

First, we need to set the function equal to zero and solve for xx.
(x)=0(x) =0x44x296=0x^{4} -4x^{2} -96 =0

STEP 3

This is a quadratic equation in the form of x2x^{2}, so we can solve it by factoring. First, we need to find two numbers that multiply to -96 and add to -. These numbers are -8 and12.
xx296=(x28)(x2+12)=0x^{} -x^{2} -96 = (x^{2} -8)(x^{2} +12) =0

STEP 4

Now, we set each factor equal to zero and solve for xx.
x28=0x^{2} -8 =0x2+12=0x^{2} +12 =0

STEP 5

olve the first equation for xx.
x2=8x^{2} =8x=±8x = \pm \sqrt{8}

STEP 6

implify the square root of8.
x=±22x = \pm2\sqrt{2}

STEP 7

olve the second equation for xx.
x2=12x^{2} = -12x=±12x = \pm \sqrt{-12}

STEP 8

Since the square root of a negative number is not a real number, there are no real solutions for the second equation. Therefore, the real zeros of the function are x=22x =2\sqrt{2} and x=22x = -2\sqrt{2}.
The zeros and the xx-intercepts are the same. They are x=22x =2\sqrt{2} and x=22x = -2\sqrt{2}.

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