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 $16-32 = -16 \Rightarrow$ $\fbox{B}$

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png