Difference between revisions of "1966 AHSME Problems/Problem 23"

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
 
<math>\fbox{A}</math>
 
<math>\fbox{A}</math>
 +
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We treat the equation as a quadratic equation in <math>y</math> for which the discriminant
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<cmath>D=16x^2-16(x+6)=16(x^2-x-6)=16(x-3)(x+2)</cmath>.
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For <math>y</math> to be real <math>D \ge 0</math>. This inequality is satisfied when <math>x \le -2</math> or <math>x \ge3</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:12, 12 December 2015

Problem

If $x$ is real and $4y^2+4xy+x+6=0$, then the complete set of values of $x$ for which $y$ is real, is:

$\text{(A) } x\le-2 \text{ or } x\ge3 \quad \text{(B) }  x\le2 \text{ or } x\ge3 \quad \text{(C) }  x\le-3 \text{ or } x\ge2 \quad \\ \text{(D) } -3\le x\le2 \quad \text{(E) } -2\le x\le3$

Solution

$\fbox{A}$

We treat the equation as a quadratic equation in $y$ for which the discriminant

\[D=16x^2-16(x+6)=16(x^2-x-6)=16(x-3)(x+2)\]. For $y$ to be real $D \ge 0$. This inequality is satisfied when $x \le -2$ or $x \ge3$.

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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