Difference between revisions of "1954 AHSME Problems/Problem 16"

(Created page with "== Problem 16== If <math>f(x) = 5x^2 - 2x - 1</math>, then <math>f(x + h) - f(x)</math> equals: <math>\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \tex...")
 
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== Solution ==
 
== Solution ==
<math>5(x+h)^2 - 2(x+h) - 1-(5x^2 - 2x - 1)\implies 5(x^2+2xh+h^2)-2x-2h-1-5x^2+2x+1\implies 10xh+5h^2-2h\implies h(10+5h-2) \boxed{(\textbf{D})}</math>
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<math>5(x+h)^2 - 2(x+h) - 1-(5x^2 - 2x - 1)\implies 5(x^2+2xh+h^2)-2x-2h-1-5x^2+2x+1\implies 10xh+5h^2-2h</math>
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<math>\implies h(10+5h-2) \boxed{(\textbf{D})}</math>
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==See Also==
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{{AHSME 50p box|year=1954|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 00:27, 28 February 2020

Problem 16

If $f(x) = 5x^2 - 2x - 1$, then $f(x + h) - f(x)$ equals:

$\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \\ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h$

Solution

$5(x+h)^2 - 2(x+h) - 1-(5x^2 - 2x - 1)\implies 5(x^2+2xh+h^2)-2x-2h-1-5x^2+2x+1\implies 10xh+5h^2-2h$

$\implies h(10+5h-2) \boxed{(\textbf{D})}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AHSME Problems and Solutions


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