Difference between revisions of "1980 AHSME Problems/Problem 24"

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== Solution ==
 
== Solution ==
<math>\fbox{}</math>
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Solution by e_power_pi_times_i
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Denote <math>s</math> as the third solution. Then, by Vieta's, <math>2r+s = \dfrac{1}{2}</math>, <math>r^2+2rs = -\dfrac{21}{4}</math>, and <math>r^2s = -\dfrac{45}{8}</math>. Multiplying the top equation by <math>2r</math> and eliminating, we have <math>3r^2 = r+\dfrac{21}{4}</math>. Combined with the fact that <math>s = \dfrac{1}{2}-2r</math>, the third equation can be written as <math>(\dfrac{r+\dfrac{21}{4}}{3})(\dfrac{1}{2}-2r) = -\dfrac{45}{8}</math>, or <math>(4r+21)(4r-1) = 135</math>. Solving, we get <math>r = \dfrac{3}{2}, -\dfrac{13}{2}</math>. Plugging the solutions back in, we see that <math>-\dfrac{13}{2}</math> is an extraneous solution, and thus the answer is <math>\boxed{\text{(D)} \ 1.52}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 13:33, 16 November 2016

Problem

For some real number $r$, the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$. Which of the following numbers is closest to $r$?

$\text{(A)} \ 1.22 \qquad  \text{(B)} \ 1.32 \qquad  \text{(C)} \ 1.42 \qquad  \text{(D)} \ 1.52 \qquad  \text{(E)} \ 1.62$

Solution

Solution by e_power_pi_times_i

Denote $s$ as the third solution. Then, by Vieta's, $2r+s = \dfrac{1}{2}$, $r^2+2rs = -\dfrac{21}{4}$, and $r^2s = -\dfrac{45}{8}$. Multiplying the top equation by $2r$ and eliminating, we have $3r^2 = r+\dfrac{21}{4}$. Combined with the fact that $s = \dfrac{1}{2}-2r$, the third equation can be written as $(\dfrac{r+\dfrac{21}{4}}{3})(\dfrac{1}{2}-2r) = -\dfrac{45}{8}$, or $(4r+21)(4r-1) = 135$. Solving, we get $r = \dfrac{3}{2}, -\dfrac{13}{2}$. Plugging the solutions back in, we see that $-\dfrac{13}{2}$ is an extraneous solution, and thus the answer is $\boxed{\text{(D)} \ 1.52}$

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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