QuestionFind the real zeros of $P(x)=x^{4}-6 x^{2}-16$ . What are the $x$ -intercepts?

Studdy Solution

STEP 1

Assumptions1. The function is $(x)=x^{4}-6 x^{}-16$ . We are looking for the real zeros of the function, which are also the $x$ -intercepts of the graph of the function.

STEP 2

To find the zeros of the function, we need to set the function equal to zero and solve for $x$ .
$x^{4}-6 x^{2}-16=0$

STEP 3

This is a quadratic equation in the form of $x^{2}$ , so we can rewrite it as $(x^{2})^{2}-6(x^{2})-16=0$ Let $y=x^{2}$ , then the equation becomes $y^{2}-6y-16=0$

STEP 4

Now, we can solve this quadratic equation using the quadratic formula $y = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ where $a=1$ , $b=-6$ , and $c=-16$ .

STEP 5

Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula to solve for $y$ .
$y = \frac{-(-) \pm \sqrt{(-)^{2}-4(1)(-16)}}{2(1)}$

STEP 6

implify the equation to find the values of $y$ .
$y = \frac{6 \pm \sqrt{36+64}}{2}$

STEP 7

Further simplify the equation to find the values of $y$ .
$y = \frac{6 \pm \sqrt{100}}{2}$

STEP 8

Calculate the square root of100.
$y = \frac{6 \pm10}{2}$

STEP 9

Calculate the two possible values of $y$ .
$y =8, -2$

STEP 10

Remember that $y=x^{2}$ , so we can substitute $y$ back in to solve for $x$ .
$x^{2} =8, -2$

STEP 11

olve for $x$ by taking the square root of both sides of the equation.
$x = \pm \sqrt{8}, \pm \sqrt{-}$

STEP 12

The square root of a negative number is not a real number, so we discard $\sqrt{-2}$ .
$x = \pm \sqrt{8}$

STEP 13

implify $\sqrt{8}$ to get the final values of $x$ .
$x = \pm2\sqrt{2}$ The zeros and the $x$ -intercepts are the same. They are $x =2\sqrt{2}, -2\sqrt{2}$ .

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