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

Latest revision as of 11:44, 5 February 2023

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

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

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

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

1954 AHSC (Problems • Answer Key • Resources)
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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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>
-<math>\implies h(10+5h-2) \boxed{(\textbf{D})}</math>
+<math>\implies h(10x+5h-2) \boxed{(\textbf{D})}</math>
 ==See Also==