Difference between revisions of "1970 AHSME Problems/Problem 14"
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<math>\text{(A) } \sqrt{4q+1}\quad | <math>\text{(A) } \sqrt{4q+1}\quad | ||
− | \text{(B) } | + | \text{(B) } q-1\quad |
\text{(C) } -\sqrt{4q+1}\quad | \text{(C) } -\sqrt{4q+1}\quad | ||
\text{(D) } q+1\quad | \text{(D) } q+1\quad | ||
Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
− | <math>\ | + | From the quadratic equation, the two roots of the equation are <math>\frac{-p\pm\sqrt{p^2-4q}}{2}</math>. The positive difference between these roots is <math>\sqrt{p^2 - 4q}</math>. Setting <math>\sqrt{p^2-4q}=1</math> and isolating <math>p</math> gives <math>\sqrt{4q+1}</math>, or choice <math>\boxed{\text{(A)}}</math>. |
== See also == | == See also == | ||
− | {{AHSME box|year=1970|num-b=13|num-a= | + | {{AHSME 35p box|year=1970|num-b=13|num-a=15}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:30, 24 February 2023
Problem
Consider , where
and
are positive numbers. If the roots of this equation differ by 1, then
equals
Solution
From the quadratic equation, the two roots of the equation are . The positive difference between these roots is
. Setting
and isolating
gives
, or choice
.
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.