University of South Carolina High School Math Contest/1993 Exam/Problem 12

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 choice must equal $3$ . The answer is $(a)$ .

Page

Toolbox

Search

University of South Carolina High School Math Contest/1993 Exam/Problem 12

Problem

Solution