Difference between revisions of "1960 AHSME Problems/Problem 39"
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==Problem== | ==Problem== | ||
+ | |||
+ | To satisfy the equation <math>\frac{a+b}{a}=\frac{b}{a+b}</math>, <math>a</math> and <math>b</math> must be: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad </math> | ||
+ | <math> \textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | First, note that <math>a \neq 0</math> and <math>a \neq -b</math>. Cross multiply both sides to get | ||
+ | <cmath>a^2 + 2ab + b^2 = ab</cmath> | ||
+ | Subtract both sides by <math>ab</math> to get | ||
+ | <cmath>a^2 + ab + b^2 = 0</cmath> | ||
+ | From the [[quadratic formula]], | ||
+ | <cmath>a = \frac{-b \pm \sqrt{b^2 - 4b^2}}{2}</cmath> | ||
+ | <cmath>a = \frac{-b \pm \sqrt{-3b^2}}{2}</cmath> | ||
+ | If <math>b</math> is [[real]], then <math>\sqrt{-3b^2}</math> is imaginary because <math>-3b^2</math> is negative. | ||
+ | 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/ZdM2ou5Gsuw?t=312 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==See Also== | ==See Also== | ||
− | {{AHSME 40p box|year=1960 | | + | {{AHSME 40p box|year=1960|num-b=38|num-a=40}} |
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 21:28, 28 December 2023
Contents
[hide]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 |