1970 AHSME Problems/Problem 6

Problem

The smallest value of $x^2+8x$ for real values of $x$ is

$\text{(A) } -16.25\quad \text{(B) } -16\quad \text{(C) } -15\quad \text{(D) } -8\quad \text{(E) None of these}$

Solution

Let's imagine this as a quadratic equation. To find the minimum or maximum value, we always need to find the vertex of the quadratic equation. The vertex of the quadratic is $\frac{-b}{2a}$ in $ax^2+bx+c=0$ . Then to find the output, or the y value of the quadratic, we plug the vertex "x" value back into the equation. In this quadratic, a=1, b=8, and c=0. So the "x" value is $\frac{-b}{2a} \Rightarrow \frac{-8}{2} = -4$ . Plugging it back into $x^2 + 8x$ , we get $16-32 = -16 \Rightarrow$ $\fbox{B}$

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1970 AHSME Problems/Problem 6

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