Difference between revisions of "1971 AHSME Problems/Problem 20"
Coolmath34 (talk | contribs) (Created page with "== Problem == The sum of the squares of the roots of the equation <math>x^2+2hx=3</math> is <math>10</math>. The absolute value of <math>h</math> is equal to <math>\textbf{(...") |
|||
| (2 intermediate revisions by one other user not shown) | |||
| Line 7: | Line 7: | ||
\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 == | + | == Solution 1 == |
| − | We can rewrite the equation as <math>x^2 + 2hx - 3 = 0.</math> By Vieta's Formulas, the sum of the roots is <math>-2h</math> and the product of the roots is <math>-3.</math> | + | We can rewrite the equation as <math>x^2 + 2hx - 3 = 0.</math> By [[Vieta's Formulas]], the sum of the roots is <math>-2h</math> and the product of the roots is <math>-3.</math> |
Let the two roots be <math>r</math> and <math>s.</math> Note that | Let the two roots be <math>r</math> and <math>s.</math> Note that | ||
<cmath>r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)</cmath> | <cmath>r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)</cmath> | ||
| − | Therefore, <math>4h^2 + 6 = 10</math> and <math>h = \pm 1.</math> This doesn't match any of the answer choices, so the answer is <math>\textbf{(E)}.</math> | + | Therefore, <math>4h^2 + 6 = 10</math> and <math>h = \pm 1.</math> This doesn't match any of the answer choices, so the answer is <math>\boxed{\textbf{(E) }\text{None of these}}.</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
| + | |||
| + | == Solution 2 == | ||
| + | The given equation can be rewritten as <math>x^2+2hx-3=0</math>. By [[Vieta's Formulas]], we know that the sum of the roots is <math>-2h</math>. 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 <math>\boxed{\textbf{(E) }\text{None of these}}</math>. | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
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. 
