1977 AHSME Problems/Problem 21: Difference between revisions

Revision as of 22:46, 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 $(B) 1$ .

Revision as of 22:45, 1 January 2024 view source Alexanderruan (talk \| contribs) editor 281 edits →Solution ← Older edit		Revision as of 22:46, 1 January 2024 view source Alexanderruan (talk \| contribs) editor 281 edits →Solution Newer edit →
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	==Solution==		==Solution==

	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)\1</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) 1</math>.

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1977 AHSME Problems/Problem 21: Difference between revisions

Revision as of 22:46, 1 January 2024

Problem 21

Solution