Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | + | From the problem, we know that | |
+ | <cmath>\frac{x+2}{6} = \frac{6}{x+2}</cmath> | ||
+ | <cmath>(x+2)^2 = 6^2</cmath> | ||
+ | <cmath>x^2+ 4x + 4 = 36</cmath> | ||
+ | <cmath>x^2 + 4x - 32 = 0</cmath> | ||
+ | <cmath>(x+8)(x-4) = 0</cmath> | ||
+ | |||
+ | Thus, <math>x = -8</math> or <math>x = 4</math>. Our answer is <math>(-8) \cdot 4=\boxed{-32}</math> | ||
+ | |||
+ | ~Bradygho | ||
+ | |||
+ | ==Solution 2== | ||
+ | We have <math>\frac{x+2}{6} = \frac{6}{x+2}</math>, so <math>x^2+4x-32=0</math>. By Vieta's our roots <math>a</math> and <math>b</math> amount to <math>\frac{-32}{1}=\boxed{-32}</math> | ||
+ | |||
+ | ~Geometry285 | ||
+ | |||
+ | ==Solution 3== | ||
+ | <math>\frac{x+2}{6}=\frac{6}{x+2} \implies x^2+4x-32</math> Therefore, the product of the root is <math>-32</math> | ||
+ | |||
+ | ~kante314 | ||
+ | |||
+ | ==Solution 4== | ||
+ | The only numbers that are their own reciprocals are <math>1</math> and <math>-1</math>. The equation <math>\frac{x+2}{6}=1</math> has the solution <math>x=4</math>, while the equation <math>\frac{x+2}{6}=-1</math> has the solution <math>x=-8</math>. The answer is <math>4 \cdot (-8)=\boxed{-32}</math>. | ||
+ | |||
+ | ~tigerzhang | ||
+ | |||
+ | |||
+ | |||
+ | ==See also== | ||
+ | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] | ||
+ | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 17:23, 11 July 2021
Problem
If is its own reciprocal, find the product of all possible values of
Solution
From the problem, we know that
Thus, or
. Our answer is
~Bradygho
Solution 2
We have , so
. By Vieta's our roots
and
amount to
~Geometry285
Solution 3
Therefore, the product of the root is
~kante314
Solution 4
The only numbers that are their own reciprocals are and
. The equation
has the solution
, while the equation
has the solution
. The answer is
.
~tigerzhang
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.