What are the solutions to the equation (x^2-6x)^2+2*(x-3)^2=81 ?

The solutions to the equation (x^2-6x)^2+2*(x-3)^2=81 are x=-1,x=3,x=7

Find the zeros of (x^2-6x)^2+2*(x-3)^2=81

The zeros of (x^2-6x)^2+2*(x-3)^2=81 are x=-1,x=3,x=7

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(x^{2} - 6 x)^{2} + 2 \cdot (x - 3)^{2} = 81

x = - 1, x = 3, x = 7

Solution steps

(x^{2} - 6 x)^{2} + 2 (x - 3)^{2} = 81

Expand

(x^{2} - 6 x)^{2} + 2 (x - 3)^{2} : x^{4} - 12 x^{3} + 38 x^{2} - 12 x + 18

x^{4} - 12 x^{3} + 38 x^{2} - 12 x + 18 = 81

Move

81

to the left side

x^{4} - 12 x^{3} + 38 x^{2} - 12 x - 63 = 0

Factor

x^{4} - 12 x^{3} + 38 x^{2} - 12 x - 63 : (x + 1) (x - 3)^{2} (x - 7)

(x + 1) (x - 3)^{2} (x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 3 = 0 or x - 7 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 3 = 0 : x = 3

Solve

x - 7 = 0 : x = 7

The solutions are

x = - 1, x = 3, x = 7

6 x^{7} - 8 = 2 (7 - 10)

(x^{2} - 8)^{2} + x^{2} - 8 = 20

x^{3} + 7 x^{2} + 17 x - 24 = 0

(x^{2} - 1) (x^{2} - 12) + 24 = 0

- (12 x^{5} - 8 x^{3}) = 25 x^{2}

What are the solutions to the equation (x^2-6x)^2+2*(x-3)^2=81 ?
The solutions to the equation (x^2-6x)^2+2*(x-3)^2=81 are x=-1,x=3,x=7
Find the zeros of (x^2-6x)^2+2*(x-3)^2=81
The zeros of (x^2-6x)^2+2*(x-3)^2=81 are x=-1,x=3,x=7