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1951 AHSME Problems/Problem 16

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Problem

If in applying the quadratic formula to a quadratic equation

f(x) \equiv ax^2 + bx + c = 0,

it happens that c = \frac{b^2}{4a}, then the graph of y = f(x) will certainly:

\mathrm{(A) \ have\ a\ maximum } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad \mathrm{(C) \ be\ tangent\ to\ the\ x-axis} \qquad \mathrm{(D) \ be\ tangent\ to\ the\ y-axis} \qquad \mathrm{(E) \ lie\ in\ one\ quadrant\ only}

Solution

The discriminant of the quadratic equation is b^2 - 4ac = b^2 - 4a\left(\frac{b^2}{4a}\right) = 0. This indicates that the equation has only one root (applying the quadratic formula, we get x = \frac{-b + \sqrt{0}}{2a} = -b/2a). Thus it follows that f(x) touches the x-axis exactly once, and hence is tangent to the x-axis \Rightarrow \mathrm{(C)}.

See also

1951 AHSME (Problems)
Preceded by
Problem 15 		Followed by
Problem 17
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
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