Difference between revisions of "1979 AHSME Problems/Problem 19"
(Created page with "== Problem 19 == Find the sum of the squares of all real numbers satisfying the equation <math>x^{256}-256^{32}=0</math>. <math>\textbf{(A) }8\qquad \textbf{(B) }128\qquad...") |
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Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
− | Notice that the solutions to the equation <math>x^{256}-1=0</math> are the <math>256</math> roots of unity. Then the solutions to the equation <math>x^{256}-256^{32}=0</math> are the <math>256</math> roots of unity dilated by <math>\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2</math>. However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are <math>\pm1</math>, and dilating by <math>2</math> gives <math>\pm2</math>. The sum of the squares is <math>(2)^2+(-2)^2 = \boxed{\textbf{(A) } 8}</math>. | + | Notice that the solutions to the equation <math>x^{256}-1=0</math> are the <math>256</math> roots of unity. Then the solutions to the equation <math>x^{256}-256^{32}=0</math> are the <math>256</math> roots of unity dilated by <math>\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2</math>. However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are <math>\pm1</math>, and dilating by <math>2</math> gives <math>\pm2</math>. The sum of the squares is <math>(2)^2+(-2)^2 = \boxed{\textbf{(A) } 8}</math> |
+ | |||
+ | ==Solution 2== | ||
+ | Notice that we can change the equation to be <math>x^{256}=256^{32}</math>. | ||
+ | |||
+ | The RHS can be simplified to <math>x^{256}=2^{256}</math> | ||
+ | |||
+ | Hence, the only real solutions are <math>x=-2,2</math> | ||
+ | |||
+ | Hence, <math>(-2)^2+(2)^2=8=\boxed{A}</math>. | ||
== See also == | == See also == |
Latest revision as of 15:09, 8 June 2017
Contents
Problem 19
Find the sum of the squares of all real numbers satisfying the equation .
Solution
Solution by e_power_pi_times_i
Notice that the solutions to the equation are the
roots of unity. Then the solutions to the equation
are the
roots of unity dilated by
. However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are
, and dilating by
gives
. The sum of the squares is
Solution 2
Notice that we can change the equation to be .
The RHS can be simplified to
Hence, the only real solutions are
Hence, .
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AHSME Problems and Solutions |
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