Difference between revisions of "1960 AHSME Problems/Problem 11"
Rockmanex3 (talk | contribs) (Created page with "==Problem== For a given value of <math>k</math> the product of the roots of <math>x^2-3kx+2k^2-1=0</math> is <math>7</math>. The roots may be characterized as: <math>\textbf...") |
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==Solution== | ==Solution== | ||
| − | If the product of the roots are <math>7</math>, then by Vieta's formulas, | + | If the product of the roots are <math>7</math>, then by [[Vieta's formulas]], |
<cmath>2k^2-1=7</cmath> | <cmath>2k^2-1=7</cmath> | ||
Solve for <math>k</math> in the resulting equation to get | Solve for <math>k</math> in the resulting equation to get | ||
| Line 14: | Line 14: | ||
<cmath>k^2=4</cmath> | <cmath>k^2=4</cmath> | ||
<cmath>k=\pm 2</cmath> | <cmath>k=\pm 2</cmath> | ||
| − | That means the two quadratics are <math>x^2-6x+7=0</math> and <math>x^2+6x+7=0</math>. Since <math>b^2</math>, <math>a</math>, and <math>c</math> are the same, the discriminant of both is | + | That means the two quadratics are <math>x^2-6x+7=0</math> and <math>x^2+6x+7=0</math>. Since <math>b^2</math>, <math>a</math>, and <math>c</math> are the same, the [[discriminant]] of both is |
<math>36-(4 \cdot 1 \cdot 7) = 8</math>. Because <math>8</math> is not a perfect square, the roots for both are irrational, so the answer is <math>\boxed{\textbf{(D)}}</math>. | <math>36-(4 \cdot 1 \cdot 7) = 8</math>. Because <math>8</math> is not a perfect square, the roots for both are irrational, so the answer is <math>\boxed{\textbf{(D)}}</math>. | ||
==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1960|num-b=10|num-a=12}} | + | {{AHSME 40p box|year=1960|num-b=10|num-a=12}} |
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 17:58, 17 May 2018
Problem
For a given value of 
the product of the roots of 
is 
. The roots may be characterized as:
Solution
If the product of the roots are , then by Vieta's formulas,
![\[2k^2-1=7\]](http://latex.artofproblemsolving.com/c/0/c/c0c3333c39193ed9c82a2c1d8614462df0f82b19.png)
Solve for in the resulting equation to get
![\[2k^2=8\]](http://latex.artofproblemsolving.com/1/c/d/1cd6df4d9d808a623f22dc841ef903a6b01cd1c1.png)
![\[k^2=4\]](http://latex.artofproblemsolving.com/4/6/4/4646a39d57b77763bb40188214fa33f2f3452f30.png)
![\[k=\pm 2\]](http://latex.artofproblemsolving.com/4/6/e/46e1269909a413132acf81c55772bd4ac7296bbd.png)
That means the two quadratics are 
and 
. Since 
, 
, and 
are the same, the discriminant of both is
. Because 
is not a perfect square, the roots for both are irrational, so the answer is 
.
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
