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Difference between revisions of "1977 AHSME Problems/Problem 21"

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Subtracting the equations, we get <math>ax+x+1+a=0</math>, or <math>(x+1)(a+1)=0</math>, so <math>x=-1</math> or <math>a=-1</math>. If <math>x=-1</math>, then <math>a=2</math>, which satisfies the condition. If <math>a=-1</math>, then <math>x</math> is nonreal. This means that <math>a=-1</math> is the only number that works, so our answer is <math>(B)</math>.
 
Subtracting the equations, we get <math>ax+x+1+a=0</math>, or <math>(x+1)(a+1)=0</math>, so <math>x=-1</math> or <math>a=-1</math>. If <math>x=-1</math>, then <math>a=2</math>, which satisfies the condition. If <math>a=-1</math>, then <math>x</math> is nonreal. This means that <math>a=-1</math> is the only number that works, so our answer is <math>(B)</math>.
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~alexanderruan

Revision as of 23:22, 1 January 2024

Problem 21

For how many values of the coefficient a do the equations \begin{align*}x^2+ax+1=0 \\ x^2-x-a=0\end{align*} have a common real solution?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \infty$

Solution

Subtracting the equations, we get $ax+x+1+a=0$, or $(x+1)(a+1)=0$, so $x=-1$ or $a=-1$. If $x=-1$, then $a=2$, which satisfies the condition. If $a=-1$, then $x$ is nonreal. This means that $a=-1$ is the only number that works, so our answer is $(B)$.

~alexanderruan

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