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Difference between revisions of "1971 AHSME Problems/Problem 20"

m (see also, boxed answer, added link)
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\textbf{(E) }\text{None of these}  </math>
 
\textbf{(E) }\text{None of these}  </math>
  
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== 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>
  
Line 18: Line 18:
  
 
-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}}

Revision as of 08:54, 5 August 2024

Problem

The sum of the squares of the roots of the equation $x^2+2hx=3$ is $10$. The absolute value of $h$ is equal to

$\textbf{(A) }-1\qquad \textbf{(B) }\textstyle\frac{1}{2}\qquad \textbf{(C) }\textstyle\frac{3}{2}\qquad \textbf{(D) }2\qquad \textbf{(E) }\text{None of these}$

Solution 1

We can rewrite the equation as $x^2 + 2hx - 3 = 0.$ By Vieta's Formulas, the sum of the roots is $-2h$ and the product of the roots is $-3.$

Let the two roots be $r$ and $s.$ Note that \[r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)\]

Therefore, $4h^2 + 6 = 10$ and $h = \pm 1.$ This doesn't match any of the answer choices, so the answer is $\boxed{\textbf{(E) }\text{None of these}}.$

-edited by coolmath34

Solution 2

The given equation can be rewritten as $x^2+2hx-3=0$. By Vieta's Formulas, we know that the sum of the roots is $-2h$. 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 $\boxed{\textbf{(E) }\text{None of these}}$.

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

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