QUESTION IMAGE
Question
find the zeros of the quadratic function, $f(x) = 2x^2 + 4x - 48$.
$x = 7$ and $x = -4$
$x = -7$ and $x = 4$
$x = 6$ and $x = -4$
$x = -6$ and $x = 4$
Step1: Set the function to zero
To find the zeros of \( f(x) = 2x^2 + 4x - 48 \), we set \( f(x) = 0 \), so we have the equation:
\[ 2x^2 + 4x - 48 = 0 \]
Step2: Simplify the equation
Divide the entire equation by 2 to simplify:
\[ \frac{2x^2 + 4x - 48}{2} = \frac{0}{2} \]
\[ x^2 + 2x - 24 = 0 \]
Step3: Factor the quadratic equation
We need to factor \( x^2 + 2x - 24 \). 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 the factored form is:
\[ (x + 6)(x - 4) = 0 \]
Step4: Solve for x
Set each factor equal to zero and solve for \( x \):
- For \( x + 6 = 0 \), we get \( x = -6 \)
- For \( x - 4 = 0 \), we get \( x = 4 \)
Wait, but let's check the options again. Wait, maybe I made a mistake in factoring? Wait, no, let's check by plugging the options into the original function.
Wait, let's check option D: \( x = -6 \) and \( x = 4 \). Let's plug \( x = -6 \) into \( f(x) = 2x^2 + 4x - 48 \):
\( f(-6) = 2(-6)^2 + 4(-6) - 48 = 2(36) - 24 - 48 = 72 - 24 - 48 = 0 \)
Plug \( x = 4 \):
\( f(4) = 2(4)^2 + 4(4) - 48 = 2(16) + 16 - 48 = 32 + 16 - 48 = 0 \)
Wait, but let's check the options. The last option is \( x = -6 \) and \( x = 4 \), which is the correct one. Wait, maybe I misread the options. Let's check the options again:
The options are:
- \( x = 7 \) and \( x = -4 \)
- \( x = -7 \) and \( x = 4 \)
- \( x = 6 \) and \( x = -4 \)
- \( x = -6 \) and \( x = 4 \)
Ah, I see, the last option is \( x = -6 \) and \( x = 4 \), which is the correct one. Let's verify by factoring again.
Original equation: \( 2x^2 + 4x - 48 = 0 \)
Divide by 2: \( x^2 + 2x - 24 = 0 \)
Factor: We need two numbers that multiply to -24 and add to 2. 6 and -4? Wait, 6 + (-4) = 2? No, 6 + (-4) = 2? Wait, 6 + (-4) = 2? 6 - 4 = 2, yes. Wait, but (x + 6)(x - 4) = x^2 - 4x + 6x - 24 = x^2 + 2x - 24, which is correct. So setting (x + 6)(x - 4) = 0 gives x = -6 and x = 4. So the correct option is the last one: \( x = -6 \) and \( x = 4 \).
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\( x = -6 \) and \( x = 4 \) (the last option, which is \( x = -6 \) and \( x = 4 \))