Difference between revisions of "1971 AHSME Problems/Problem 20"
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\textbf{(C) }\textstyle\frac{3}{2}\qquad | \textbf{(C) }\textstyle\frac{3}{2}\qquad | ||
\textbf{(D) }2\qquad | \textbf{(D) }2\qquad | ||
| − | \textbf{(E) }\text{None of these} | + | \textbf{(E) }\text{None of these}</math> |
== Solution 1 == | == Solution 1 == | ||
Latest revision as of 13:41, 23 September 2024
Contents
Problem
The sum of the squares of the roots of the equation 
is 
. The absolute value of 
is equal to
Solution 1
We can rewrite the equation as 
By Vieta's Formulas, the sum of the roots is 
and the product of the roots is 
Let the two roots be 
and 
Note that
![\[r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)\]](http://latex.artofproblemsolving.com/b/9/9/b99eba5478869ca6bcde048172e6a2fe4367680d.png)
Therefore, 
and 
This doesn't match any of the answer choices, so the answer is 
-edited by coolmath34
Solution 2
The given equation can be rewritten as 
. By Vieta's Formulas, we know that the sum of the roots is . Thus, by Newton Sums, we have the following equation:
\begin{align*}
10+(2h)(-2h)+2(-3) &= 0 \\
10-4h^2-6 &= 0 \\
-4h^2 &= -4 \\
h^2 &= 1 \\
|h| &= 1 \\
\end{align*}
Thus, our answer is 
.
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
