Difference between revisions of "1988 AHSME Problems/Problem 5"
(Created page with "==Problem== If <math>b</math> and <math>c</math> are constants and <math>(x + 2)(x + b) = x^2 + cx + 6</math>, then <math>c</math> is <math>\textbf{(A)}\ -5\qquad \textbf{(B)}\...") |
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==Solution== | ==Solution== | ||
| − | + | We first start out by expanding the left side of the equation, <math>(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6</math>. We know the constant | |
| − | + | terms have to be equal so we have <math>2b=6</math>, so <math>b=3</math>. Plugging <math>b=3</math> back in yields <math>x^2+(2+3)x+6=x^2+cx+6</math>. Thus, <math>c=5 \implies \boxed{\text{E}}</math>. | |
== See also == | == See also == | ||
Latest revision as of 05:37, 31 August 2015
Problem
If 
and 
are constants and 
, then is
Solution
We first start out by expanding the left side of the equation, 
. We know the constant
terms have to be equal so we have 
, so 
. Plugging back in yields 
. Thus, 
.
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
