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 == | ||
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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
