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

(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{(...")
 
m (see also, boxed answer, added link)
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== Solution ==
 
== Solution ==
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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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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
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 +
== See Also ==
 +
{{AHSME 35p box|year=1971|num-b=19|num-a=21}}
 +
{{MAA Notice}}

Revision as of 09:49, 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

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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