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==
 
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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
 
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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 06:37, 31 August 2015

Problem

If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$, then $c$ is

$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$


Solution

We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$. We know the constant terms have to be equal so we have $2b=6$, so $b=3$. Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$. Thus, $c=5 \implies \boxed{\text{E}}$.

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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