Difference between revisions of "1957 AHSME Problems/Problem 45"
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== Solution == | == Solution == | ||
− | <math>\boxed{\textbf{(A)}}</math>. | + | From the initial equation <math>\tfrac x y = x-y</math>, we can solve for <math>y</math> in terms of <math>x</math> as follows: |
+ | \begin{align*}\ | ||
+ | \frac x y &= x-y \ | ||
+ | x &= xy-y^2 \ | ||
+ | y^2-xy+x &= 0 \ | ||
+ | y = \frac{x \pm \sqrt{x^2-4x}}2 | ||
+ | \end{align*} | ||
+ | Because <math>y</math> is real, the [[discriminant]] of the above expression for <math>y</math> must be <math>\geq 0</math>, so <math>x^2-4x=x(x-4) \geq 0</math>. This is true when <math>x \geq 4</math> and <math>x \leq 0</math>, so we choose answer <math>\boxed{\textbf{(A)}}</math>. | ||
== See Also == | == See Also == |
Latest revision as of 11:06, 27 July 2024
Problem
If two real numbers and satisfy the equation , then:
Solution
From the initial equation , we can solve for in terms of as follows:
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 44 |
Followed by Problem 46 | |
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