Difference between revisions of "1975 AHSME Problems/Problem 12"

m (Solution: Fixed formatting on the answer)
 
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<cmath>ab=6x^2.</cmath>
 
<cmath>ab=6x^2.</cmath>
 
Checking with the possible answers, along with <math>a-b=x</math> yields the only answer to be <math>\boxed{(\textbf{B}): a=3x\text{ or }a=-2x}.</math>
 
Checking with the possible answers, along with <math>a-b=x</math> yields the only answer to be <math>\boxed{(\textbf{B}): a=3x\text{ or }a=-2x}.</math>
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==See Also==
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{{AHSME box|year=1975|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 16:09, 19 January 2021

Problem 12

If $a \neq b, a^3 - b^3 = 19x^3$, and $a-b = x$, which of the following conclusions is correct?

$\textbf{(A)}\ a=3x \qquad  \textbf{(B)}\ a=3x \text{ or } a = -2x \qquad  \textbf{(C)}\ a=-3x \text{ or } a = 2x \qquad \\ \textbf{(D)}\ a=3x \text{ or } a=2x \qquad  \textbf{(E)}\ a=2x$

Solution

We can factor $a^3-b^3=19x^3$ into: \[(a-b)(a^2+ab+b^2)=19x^3.\] Substituting yields: \[x(a^2+ab+b^2)=19x^3\] \[a^2+ab+b^2=19x^2.\] This is equal to: \[(a-b)^2+3ab=19x^2\] \[x^2+3ab=19x^2\] \[ab=6x^2.\] Checking with the possible answers, along with $a-b=x$ yields the only answer to be $\boxed{(\textbf{B}): a=3x\text{ or }a=-2x}.$

See Also

1975 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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