Asked by maritza
how do you solve this problem (m-5)^4+36=13(m-5)^2 algebraically
Answers
Answered by
Reiny
(m-5)^4 + 36 = (m-5)^2
let (m-5)^2 = x, so your equation becomes
x^2 + 36 = x
x^2 - x + 36 = 0
x^2 = (1 ± √(1-144) )/2
= (1 ± √-143)/2
then (m - 5)^2 = (1 ± √-143)/2
m-5 = ±√(1 ± √-143)/2
m = 5 ±√(1 ± √-143)/2
= [ 10 ±√(1 ± √-143) ]/2
or as Wolfram has it:
(1/2) [ 10 ±√(1 ± √-143) ]
or
(1/2) [ 10 ±√(1 ± i√143) ]
http://www.wolframalpha.com/input/?i=solve+%28m-5%29%5E4+%2B+36+%3D+%28m-5%29%5E2
let (m-5)^2 = x, so your equation becomes
x^2 + 36 = x
x^2 - x + 36 = 0
x^2 = (1 ± √(1-144) )/2
= (1 ± √-143)/2
then (m - 5)^2 = (1 ± √-143)/2
m-5 = ±√(1 ± √-143)/2
m = 5 ±√(1 ± √-143)/2
= [ 10 ±√(1 ± √-143) ]/2
or as Wolfram has it:
(1/2) [ 10 ±√(1 ± √-143) ]
or
(1/2) [ 10 ±√(1 ± i√143) ]
http://www.wolframalpha.com/input/?i=solve+%28m-5%29%5E4+%2B+36+%3D+%28m-5%29%5E2
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