1953 AHSME Problems/Problem 36: Difference between revisions

Latest revision as of 00:41, 4 February 2020

Problem

Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$ . The obtained value, $m$ , is an exact divisor of:

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$

Solution

Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$ , where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$ . The only number that fits the first equation is $y=6$ , so $m=-18$ . The only choice that is a multiple of 18 is $\boxed{\textbf{(C) }36}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 35	Followed by Problem 37
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All AHSME Problems and Solutions

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 ==Solution==
-Since the given expression is a quadratic, the factored form would be <math>(x-3)(4x+y)</math>, where <math>y</math> is a value such that <math>-12x+yx=-6x</math> and <math>-3(y)=m</math>. The only number that fits the first equation is <math>y=6</math>, so <math>m=18</math>. The only choice that is a multiple of 18 is <math>\boxed{\textbf{(C) }36}</math>.
+Since the given expression is a quadratic, the factored form would be <math>(x-3)(4x+y)</math>, where <math>y</math> is a value such that <math>-12x+yx=-6x</math> and <math>-3(y)=m</math>. The only number that fits the first equation is <math>y=6</math>, so <math>m=-18</math>. The only choice that is a multiple of 18 is <math>\boxed{\textbf{(C) }36}</math>.
+==See Also==
+{{AHSME 50p box|year=1953|num-b=35|num-a=37}}
+{{MAA Notice}}

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1953 AHSME Problems/Problem 36: Difference between revisions

Latest revision as of 00:41, 4 February 2020

Problem

Solution

See Also