1969 AHSME Problems/Problem 12

Revision as of 03:20, 7 June 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 12)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Let $F=\frac{6x^2+16x+3m}{6}$ be the square of an expression which is linear in $x$. Then $m$ has a particular value between:

$\text{(A) } 3 \text{ and } 4\quad \text{(B) } 4 \text{ and } 5\quad \text{(C) } 5 \text{ and } 6\quad \text{(D) } -4 \text{ and } -3\quad \text{(E) } -6 \text{ and } -5$

Solution

The quadratic can also be written as \[x^2 + \frac{8}{3}x + \frac{m}{2}\] In order for the quadratic to be a square of a linear expression, the constant term must be the square of the x-term divided by 4. Thus, \[\frac{m}{2} = (\frac{8}{3})^2 \cdot \frac{1}{4}\] \[\frac{m}{2} = \frac{16}{9}\] \[m = \frac{32}{9}\] The answer is $\boxed{\textbf{(A)}}$.

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png