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

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\text{(E)} \ 1.62  </math>   
 
\text{(E)} \ 1.62  </math>   
  
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== Solution ==
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== Solution 1 ==
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Solution by e_power_pi_times_i
 
  
 
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>
 
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>
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-e_power_pi_times_i
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== Solution 2 ==
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By polynomial long division, <math>\frac{8x^3 - 4x^2 - 42x + 45}{x^2 - 2rx + r^2} = 8x + (16r-4) + \frac{(24r^2 - 8r - 42)x + (-16r^3 + 4r^2 + 45)}{x^2 - 2rx + r^2}</math>.
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So <math>(24r^2 - 8r - 42)x + (-16r^3 + 4r^2 + 45) = 0</math>, that is, <math>r</math> is a root of both <math>24r^2 - 8r - 42</math> and <math>-16r^3 + 4r^2 + 45</math>.
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Since <math>24r^2 - 8r - 42 = 2(6r + 7)(2r - 3)</math>, either <math>r = -\frac{7}{6}</math> or <math>r = \frac{3}{2}</math>.  By the Rational Root Theorem, <math>-\frac{7}{6}</math> is not a root of <math>-16r^3 + 4r^2 + 45</math>.  However, <math>-16\left(\frac{3}{2}\right)^3 + 4\left(\frac{3}{2}\right)^2 + 45 = -16 \cdot \frac{27}{8} + 4 \cdot \frac{9}{4} + 45 = -54 + 9 + 45 = 0</math>, so <math>\frac{3}{2}</math> is a root of <math>-16r^3 + 4r^2 + 45</math>.
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Therefore, <math>r = \frac{3}{2} = 1.5</math>, so <math>\boxed{(\textbf{D})\ 1.52}</math> is closest to <math>r</math>.
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-j314andrews
  
 
== See also ==
 
== See also ==

Revision as of 18:24, 16 August 2025

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 1

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 $\boxed{\text{(D)} \ 1.52}$

-e_power_pi_times_i

Solution 2

By polynomial long division, $\frac{8x^3 - 4x^2 - 42x + 45}{x^2 - 2rx + r^2} = 8x + (16r-4) + \frac{(24r^2 - 8r - 42)x + (-16r^3 + 4r^2 + 45)}{x^2 - 2rx + r^2}$.

So $(24r^2 - 8r - 42)x + (-16r^3 + 4r^2 + 45) = 0$, that is, $r$ is a root of both $24r^2 - 8r - 42$ and $-16r^3 + 4r^2 + 45$.

Since $24r^2 - 8r - 42 = 2(6r + 7)(2r - 3)$, either $r = -\frac{7}{6}$ or $r = \frac{3}{2}$. By the Rational Root Theorem, $-\frac{7}{6}$ is not a root of $-16r^3 + 4r^2 + 45$. However, $-16\left(\frac{3}{2}\right)^3 + 4\left(\frac{3}{2}\right)^2 + 45 = -16 \cdot \frac{27}{8} + 4 \cdot \frac{9}{4} + 45 = -54 + 9 + 45 = 0$, so $\frac{3}{2}$ is a root of $-16r^3 + 4r^2 + 45$.

Therefore, $r = \frac{3}{2} = 1.5$, so $\boxed{(\textbf{D})\ 1.52}$ is closest to $r$.

-j314andrews

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AHSME Problems and Solutions

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