1967 AHSME Problems/Problem 17: Difference between revisions
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 == | ||
{{AHSME box|year=1967|num-b=16|num-a=18}} | {{AHSME 40p box|year=1967|num-b=16|num-a=18}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:40, 16 August 2023
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 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
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