Solution set of the equation (x−1)^4+(x−5)^4= 82 is

Answer 1

(x-1)^4 + (x-5)^4 = 82

(x^2 -2x +1)^2 + (x^2 -10x +25)^2 = 82

2x^4 -24x^3 +155x^2 -504x +626 = 82

by decartes rule max 4 real positive solutions and no negative real zeros

any 4th degree equation has 4 solutions, although some may repeat or be imaginary

x = 2 and x =4

(2-1)^4 + (2-5)^4

= 1 + 81 = 82

(4-1)^4 + (4-5)^4

= 81 + 1 = 82

possibly 2 more, but likely 2 other imaginary zeros (they come in conjugate pairs)

real solution set: {2,4}

graph the equation. It has two x intercepts, 2 and 4

which means 2 imaginary solutions

(x-1)^4 + (x-5)^2 = 82

(x^2 -2x +1)^2 + (x^2 -10x +25)^2= (sqr82)^2

looks like the equation for a circle, but it's not

x=3+5i

x-3 = 5i

x^2 -6x + 9= -25

x^2 -6x +34 = 0

left side is one factor

(x-2)(x-4)

= x^2 -6x +8 = 0

left side is a 2nd factor

multiply them together

(x^2 -6x+8)(x^2 -6x+34) = 0

x^4 -12x^3-252x+272=0

2x^4-24x^3-504x +544=0

2x^4-24x^3-504x+626= 82

(x^2-2x+1)^2 +(x^2-10x+25)^2 = 82

(x-1)^4 +(x-5)^4 = 82