Question

question 1-15
find the zeros of the quadratic function, $f(x) = 2x^2 + 4x - 48$.
\\(\circ\\) $x = 7$ and $x = -4$
\\(\circ\\) $x = -7$ and $x = 4$
\\(\circ\\) $x = 6$ and $x = -4$
\\(\circ\\) $x = -6$ and $x = 4$

Explanation:

Step1: Set \( f(x) = 0 \)

To find the zeros of the quadratic function \( f(x)=2x^{2}+4x - 48 \), we start by setting \( f(x) = 0 \). So we have the equation:
\[
2x^{2}+4x - 48=0
\]

Step2: Simplify the equation

We can divide the entire equation by 2 to simplify it. Dividing each term by 2 gives:
\[
x^{2}+2x - 24 = 0
\]

Step3: Factor the quadratic equation

We need to factor the quadratic \( x^{2}+2x - 24 \). We look for two numbers that multiply to - 24 and add up to 2. The numbers are 6 and - 4 because \( 6\times(-4)=-24 \) and \( 6+(-4) = 2 \). So we can factor the quadratic as:
\[
(x + 6)(x - 4)=0
\]

Step4: Solve for \( x \)

Using the zero - product property, if \( (x + 6)(x - 4)=0 \), then either \( x+6 = 0 \) or \( x - 4=0 \).

• If \( x+6 = 0 \), then \( x=-6 \).
• If \( x - 4=0 \), then \( x = 4 \).

Answer:

\( x=-6 \) and \( x = 4 \) (corresponding to the option: \( x=-6 \) and \( x = 4 \))