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Difference between revisions of "1958 AHSME Problems/Problem 39"

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Note that for all roots <math>x</math>, <math>-x</math> will also be a root. Therefore, the sum of all of the roots will be <math>0</math>, making the answer <math>\fbox{C}</math>
 
Note that for all roots <math>x</math>, <math>-x</math> will also be a root. Therefore, the sum of all of the roots will be <math>0</math>, making the answer <math>\fbox{C}</math>
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We can find all roots <math>x</math> by setting <math>|x| = y</math>. This gives us the equation <math>y^2+y-6=0</math>, which has the solutions <math>y=-3, 2</math>. However, <math>|x|</math> cannot equal <math>-3</math>, so the roots for <math>x</math> are <math>2</math> and <math>-2</math>. The sum of the two roots is <math>0</math>, making the answer <math>\fbox{C}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:24, 22 February 2018

Problem

We may say concerning the solution of $|x|^2 + |x| - 6 =0$ that:

$\textbf{(A)}\ \text{there is only one root}\qquad \textbf{(B)}\ \text{the sum of the roots is }{+1}\qquad \textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \\ \textbf{(D)}\ \text{the product of the roots is }{+4}\qquad \textbf{(E)}\ \text{the product of the roots is }{-6}$

Solution

Note that for all roots $x$, $-x$ will also be a root. Therefore, the sum of all of the roots will be $0$, making the answer $\fbox{C}$


We can find all roots $x$ by setting $|x| = y$. This gives us the equation $y^2+y-6=0$, which has the solutions $y=-3, 2$. However, $|x|$ cannot equal $-3$, so the roots for $x$ are $2$ and $-2$. The sum of the two roots is $0$, making the answer $\fbox{C}$.

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 38 		Followed by
Problem 40
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All AHSME Problems and Solutions

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