Difference between revisions of "1971 AHSME Problems/Problem 20"
m (see also, boxed answer, added link) |
(added second solution) |
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\textbf{(E) }\text{None of these} </math> | \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> | ||
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-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 == | == See Also == | ||
{{AHSME 35p box|year=1971|num-b=19|num-a=21}} | {{AHSME 35p box|year=1971|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:54, 5 August 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
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.