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Difference between revisions of "1962 AHSME Problems/Problem 34"

(Created page with "==Problem== For what real values of <math>K</math> does <math>x = K^2 (x-1)(x-2)</math> have real roots? <math> \textbf{(A)}\ \text{none}\qquad\textbf{(B)}\ -2<K<1\qquad\textbf...")
 
 
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==Solution==
 
==Solution==
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"Unsolved"
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Factoring, we get
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<cmath>x = K^2x^2 - 3K^2x + 2K^2</cmath>
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<cmath>K^2x^2 - (3K^2+1)x + 2K^2 = 0</cmath>
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At this point, we could use the quadratic formula, but we're only interested in whether the roots are real, which occurs if and only if the discriminant (<math>b^2-4ac</math>) is non-negative.
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<cmath>(-3K^2-1)^2-8K^4\ge0</cmath>
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<cmath>9K^4+6K^2+1-8K^4\ge0</cmath>
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<cmath>K^4+6K^2+1\ge0</cmath>
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This function of <math>K</math> is symmetric with respect to the y-axis. Plugging in <math>0</math> for <math>K</math> gives us <math>1</math>, and it is clear that the function is strictly increasing over <math>(0,\infty)</math>, and so it is positive for all real values of <math>K</math>. <math>\boxed{\textbf{(E)}}</math>

Latest revision as of 16:37, 17 April 2014

Problem

For what real values of $K$ does $x = K^2 (x-1)(x-2)$ have real roots?

$\textbf{(A)}\ \text{none}\qquad\textbf{(B)}\ -2<K<1\qquad\textbf{(C)}\ -2\sqrt{2}< K < 2\sqrt{2}\qquad$

$\textbf{(D)}\ K>1\text{ or }K<-2\qquad\textbf{(E)}\ \text{all}$

Solution

Factoring, we get \[x = K^2x^2 - 3K^2x + 2K^2\] \[K^2x^2 - (3K^2+1)x + 2K^2 = 0\] At this point, we could use the quadratic formula, but we're only interested in whether the roots are real, which occurs if and only if the discriminant ($b^2-4ac$) is non-negative. \[(-3K^2-1)^2-8K^4\ge0\] \[9K^4+6K^2+1-8K^4\ge0\] \[K^4+6K^2+1\ge0\] This function of $K$ is symmetric with respect to the y-axis. Plugging in $0$ for $K$ gives us $1$, and it is clear that the function is strictly increasing over $(0,\infty)$, and so it is positive for all real values of $K$. $\boxed{\textbf{(E)}}$

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