What are the solutions to the equation (t+3)^4+(t-3)^4=272 ?

The solutions to the equation (t+3)^4+(t-3)^4=272 are t=1,t=-1,t=sqrt(55)i,t=-sqrt(55)i

Find the zeros of (t+3)^4+(t-3)^4=272

The zeros of (t+3)^4+(t-3)^4=272 are t=1,t=-1,t=sqrt(55)i,t=-sqrt(55)i

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(t + 3)^{4} + (t - 3)^{4} = 272

t = 1, t = - 1, t = 55 i, t = - 55 i

Solution steps

(t + 3)^{4} + (t - 3)^{4} = 272

Expand

(t + 3)^{4} + (t - 3)^{4} : 2 t^{4} + 108 t^{2} + 162

2 t^{4} + 108 t^{2} + 162 = 272

Move

272

to the left side

2 t^{4} + 108 t^{2} - 110 = 0

Rewrite the equation with

u = t^{2}

and

u^{2} = t^{4}

2 u^{2} + 108 u - 110 = 0

Solve

2 u^{2} + 108 u - 110 = 0 : u = 1, u = - 55

u = 1, u = - 55

Substitute back

u = t^{2},

solve for

t

Solve

t^{2} = 1 : t = 1, t = - 1

Solve

t^{2} = - 55 : t = 55 i, t = - 55 i

The solutions are

t = 1, t = - 1, t = 55 i, t = - 55 i

(x + 2) + (x^{2} - 4 x + 4) - x^{2} (x - 2) = - 53

64 x^{3} + 1 = 0

x^{6} + 6 x^{3} + 9 = 0

d^{3} - 6 d^{2} + 11 d - 6 = 0

x^{100} = 1

What are the solutions to the equation (t+3)^4+(t-3)^4=272 ?
The solutions to the equation (t+3)^4+(t-3)^4=272 are t=1,t=-1,t=sqrt(55)i,t=-sqrt(55)i
Find the zeros of (t+3)^4+(t-3)^4=272
The zeros of (t+3)^4+(t-3)^4=272 are t=1,t=-1,t=sqrt(55)i,t=-sqrt(55)i