1960 AHSME Problems/Problem 39: Difference between revisions
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If <math>b</math> is not real, where <math>b = m+ni</math> and <math>n \neq 0</math>, then <math>\sqrt{-3b^2}</math> evaluates to <math>\sqrt{-3m^2 - 6mni + 3n^2}</math>. As long as <math>m \neq 0</math>, the expression can also be imaginary because a real number squared will be a real number. | If <math>b</math> is not real, where <math>b = m+ni</math> and <math>n \neq 0</math>, then <math>\sqrt{-3b^2}</math> evaluates to <math>\sqrt{-3m^2 - 6mni + 3n^2}</math>. As long as <math>m \neq 0</math>, the expression can also be imaginary because a real number squared will be a real number. | ||
From these two points, the answer is <math>\boxed{\textbf{(E)}}</math>. | From these two points, the answer is <math>\boxed{\textbf{(E)}}</math>. | ||
==Video Solution== | |||
https://youtu.be/ZdM2ou5Gsuw?t=312 | |||
~MathProblemSolvingSkills.com | |||
==See Also== | ==See Also== | ||
Latest revision as of 21:28, 28 December 2023
Problem
To satisfy the equation
,
and
must be:
Solution
First, note that
and
. Cross multiply both sides to get
Subtract both sides by
to get
From the quadratic formula,
If
is real, then
is imaginary because
is negative.
If
is not real, where
and
, then
evaluates to
. As long as
, the expression can also be imaginary because a real number squared will be a real number.
From these two points, the answer is
.
Video Solution
https://youtu.be/ZdM2ou5Gsuw?t=312
~MathProblemSolvingSkills.com
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||