Question

solving quadratic equations: zero product property

which function has real zeros at \\(x = -10\\) and \\(x = -6\\)?

\\(f(x) = x^2 + 4x + 60\\)

\\(f(x) = x^2 + 16x + 60\\)

\\(f(x) = x^2 - 16x + 60\\)

Explanation:

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<reasoning>

Set up the factored form of the quadratic function

\[

$$\begin{aligned} x_1 &= -10 \\ x_2 &= -6 \\ f(x) &= a(x - x_1)(x - x_2) \\ f(x) &= a(x - (-10))(x - (-6)) \\ f(x) &= a(x + 10)(x + 6) \end{aligned}$$

\]

Expand the factors to standard form

\[

$$\begin{aligned} a &= 1 \\ f(x) &= (x + 10)(x + 6) \\ f(x) &= x^2 + 6x + 10x + 60 \\ f(x) &= x^2 + 16x + 60 \end{aligned}$$

\]

Match with the given options

\[

$$\begin{aligned} f(x) &= x^2 + 16x + 60 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(f(x) = x^2 + 4x + 60\)</mcq-option>
<mcq-correct>(B) \(f(x) = x^2 + 16x + 60\)</mcq-correct>
<mcq-option>(C) \(f(x) = x^2 - 16x + 60\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Set up the factored form of the quadratic function

\[

$$\begin{aligned} x_1 &= -10 \\ x_2 &= -6 \\ f(x) &= a(x - x_1)(x - x_2) \\ f(x) &= a(x - (-10))(x - (-6)) \\ f(x) &= a(x + 10)(x + 6) \end{aligned}$$

\]

Expand the factors to standard form

\[

$$\begin{aligned} a &= 1 \\ f(x) &= (x + 10)(x + 6) \\ f(x) &= x^2 + 6x + 10x + 60 \\ f(x) &= x^2 + 16x + 60 \end{aligned}$$

\]

Match with the given options

\[

$$\begin{aligned} f(x) &= x^2 + 16x + 60 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(f(x) = x^2 + 4x + 60\)</mcq-option>
<mcq-correct>(B) \(f(x) = x^2 + 16x + 60\)</mcq-correct>
<mcq-option>(C) \(f(x) = x^2 - 16x + 60\)</mcq-option>
</answer>

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