Difference between revisions of "1966 AHSME Problems/Problem 23"
(Created page with "== Problem == If <math>x</math> is real and <math>4y^2+4xy+x+6=0</math>, then the complete set of values of <math>x</math> for which <math>y</math> is real, is: <math>\text{(A) ...") |
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== Solution == | == Solution == | ||
| + | We treat the equation as a quadratic equation in <math>y</math> for which the discriminant | ||
| + | <cmath>D=16x^2-16(x+6)=16(x^2-x-6)=16(x-3)(x+2)</cmath> | ||
| + | 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 
is real and 
, then the complete set of values of for which 
is real, is:
Solution
We treat the equation as a quadratic equation in for which the discriminant
![\[D=16x^2-16(x+6)=16(x^2-x-6)=16(x-3)(x+2)\]](http://latex.artofproblemsolving.com/8/5/5/85505f1639cbb612a29ee8785680a2e373ffc556.png)
For to be real 
. This inequality is satisfied when 
or 
or 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
