Difference between revisions of "1967 AHSME Problems/Problem 17"
(Created page with "== Problem == If <math>r_1</math> and <math>r_2</math> are the distinct real roots of <math>x^2+px+8=0</math>, then it must follow that: <math>\textbf{(A)}\ |r_1+r_2|>4\sqrt{2}\...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{A}</math> | + | We are given that the roots are real, so the discriminant is positive, which means <math>p^2 - 4(8)(1) > 0</math>. This leads to <math>|p| > 4\sqrt{2}</math>. By Vieta, the sum of the roots is <math>-p</math>, so we have <math>|-(r_1 + r_2)| \ge 4\sqrt{2}</math>, or <math>|r_1 + r_2| > 4\sqrt{2}</math>, which is option <math>\fbox{A}</math>. |
== See also == | == See also == |
Latest revision as of 02:03, 13 July 2019
Problem
If and are the distinct real roots of , then it must follow that:
Solution
We are given that the roots are real, so the discriminant is positive, which means . This leads to . By Vieta, the sum of the roots is , so we have , or , which is option .
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AHSME Problems and Solutions |
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