Difference between revisions of "1951 AHSME Problems/Problem 16"

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The [[discriminant]] of the quadratic equation is <math>b^2 - 4ac = b^2 - 4a\left(\frac{b^2}{4a}\right) = 0</math>. This indicates that the equation has only one root (applying the quadratic formula, we get <math>x = \frac{-b + \sqrt{0}}{2a} = -b/2a</math>). Thus it follows that <math>f(x)</math> touches the x-axis exactly once, and hence is tangent to the x-axis <math>\Rightarrow \mathrm{(C)}</math>.
 
The [[discriminant]] of the quadratic equation is <math>b^2 - 4ac = b^2 - 4a\left(\frac{b^2}{4a}\right) = 0</math>. This indicates that the equation has only one root (applying the quadratic formula, we get <math>x = \frac{-b + \sqrt{0}}{2a} = -b/2a</math>). Thus it follows that <math>f(x)</math> touches the x-axis exactly once, and hence is tangent to the x-axis <math>\Rightarrow \mathrm{(C)}</math>.
  
== See also ==
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== See Also ==
{{AHSME box|year=1951|num-b=15|num-a=17}}  
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{{AHSME 50p box|year=1951|num-b=15|num-a=17}}  
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 07:55, 29 April 2012

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 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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