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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 12"

 
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== Problem ==
 
== Problem ==
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If the equations <math> (1) x^2 + ax + b = 0</math> and <math> (2) x^2 + cx + d = 0 </math> have exactly one root in common, and <math> abcd\ne 0,</math> then the other root of equation <math> (2) </math> is
  
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<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
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<center><math> \mathrm{(A) \ }\frac{c-a}{b-d}d \qquad \mathrm{(B) \ }\frac{a+c}{b+d}d \qquad \mathrm{(C) \ }\frac{b+c}{a+d}c \qquad \mathrm{(D) \ }\frac{a-c}{b-d} \qquad \mathrm{(E) \ }\frac{a+c}{b-d}c </math></center>
  
 
== Solution ==
 
== Solution ==
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Let <math>(1)</math> have roots <math>x=1,2</math> and <math>(2)</math> have roots <math>x=1,3</math>. Thus:
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<math>(1)</math> <math>x^{2}-3x+2=0</math>
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<math>(2)</math> <math>x^{2}-4x+3=0</math>
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Thus, we know that <math>(a,b,c,d)=(-3,2,-4,3)</math> and our answer coice must equal <math>3</math>. The answer is <math>(a)</math>.
  
 
== See also ==
 
== See also ==
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
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 +
[[Category:Intermediate Algebra Problems]]

Revision as of 20:31, 22 July 2006

Problem

If the equations $(1) x^2 + ax + b = 0$ and $(2) x^2 + cx + d = 0$ have exactly one root in common, and $abcd\ne 0,$ then the other root of equation $(2)$ is

$\mathrm{(A) \ }\frac{c-a}{b-d}d \qquad \mathrm{(B) \ }\frac{a+c}{b+d}d \qquad \mathrm{(C) \ }\frac{b+c}{a+d}c \qquad \mathrm{(D) \ }\frac{a-c}{b-d} \qquad \mathrm{(E) \ }\frac{a+c}{b-d}c$

Solution

Let $(1)$ have roots $x=1,2$ and $(2)$ have roots $x=1,3$. Thus: $(1)$ $x^{2}-3x+2=0$ $(2)$ $x^{2}-4x+3=0$

Thus, we know that $(a,b,c,d)=(-3,2,-4,3)$ and our answer coice must equal $3$. The answer is $(a)$.

See also

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