Difference between revisions of "1980 AHSME Problems/Problem 24"
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== Solution == | == Solution == | ||
− | <math>\ | + | 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 , the polynomial is divisible by . Which of the following numbers is closest to ?
Solution
Solution by e_power_pi_times_i
Denote as the third solution. Then, by Vieta's, , , and . Multiplying the top equation by and eliminating, we have . Combined with the fact that , the third equation can be written as , or . Solving, we get . Plugging the solutions back in, we see that is an extraneous solution, and thus the answer is
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AHSME Problems and Solutions |
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