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

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== Solution ==
 
== Solution ==
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<math>\fbox{A}</math>
 
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We treat the equation as a quadratic equation in <math>y</math> for which the discriminant  
 
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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<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>.
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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> or <math>\fbox{A}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 22:13, 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

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 $x \ge3$ or $\fbox{A}$

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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