QuestionFind the real zeros of the function $f(x)=x^{2}+6x+4$ using the quadratic formula. Choose A, B, or C for answers.

Studdy Solution

STEP 1

Assumptions1. The function is a quadratic function, given by $f(x)=x^{}+6 x+4$ . We are asked to find the real zeros of the function, if any, using the quadratic formula.
3. The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^ -4ac}}{a}$ , where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation $ax^ + bx + c =0$ .

STEP 2

First, we need to identify the coefficients $a$ , $b$ , and $c$ from the given quadratic function.From $f(x)=x^{2}+6 x+4$ , we can see that $a =1$ , $b =6$ , and $c =4$

STEP 3

Now, we substitute these values into the quadratic formula to find the solutions for $x$ .
$x = \frac{-b \pm \sqrt{b^2 -ac}}{2a}$

STEP 4

Substitute $a =1$ , $b =6$ , and $c =4$ into the formula.
$x = \frac{-6 \pm \sqrt{6^2 -414}}{2*1}$

STEP 5

implify the expression under the square root.
$x = \frac{- \pm \sqrt{36 -16}}{2}$

STEP 6

Calculate the value under the square root.
$x = \frac{-6 \pm \sqrt{20}}{2}$

STEP 7

implify the square root. $\sqrt{20}$ can be simplified to $2\sqrt{5}$ .
$x = \frac{-6 \pm2\sqrt{5}}{2}$

STEP 8

Now, divide each term in the numerator by2.
$x = -3 \pm \sqrt{5}$ These are the real zeros of the function $f(x)=x^{2}+6 x+4$ .So, the correct choice is A. The zeros and the $x$ -intercepts are the same. They are $-3 + \sqrt{5}$ and $-3 - \sqrt{5}$ .

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