What are the solutions to the equation x^4=(4x-5)^2 ?

The solutions to the equation x^4=(4x-5)^2 are x=1,x=-5,x=2+i,x=2-i

Find the zeros of x^4=(4x-5)^2

The zeros of x^4=(4x-5)^2 are x=1,x=-5,x=2+i,x=2-i

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x^{4} = (4 x - 5)^{2}

x = 1, x = - 5, x = 2 + i, x = 2 - i

Solution steps

x^{4} = (4 x - 5)^{2}

Expand

(4 x - 5)^{2} : 16 x^{2} - 40 x + 25

x^{4} = 16 x^{2} - 40 x + 25

Move

25

to the left side

x^{4} - 25 = 16 x^{2} - 40 x

Move

40 x

to the left side

x^{4} - 25 + 40 x = 16 x^{2}

Move

16 x^{2}

to the left side

x^{4} - 25 + 40 x - 16 x^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{4} - 16 x^{2} + 40 x - 25 = 0

Factor

x^{4} - 16 x^{2} + 40 x - 25 : (x - 1) (x + 5) (x^{2} - 4 x + 5)

(x - 1) (x + 5) (x^{2} - 4 x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 5 = 0 or x^{2} - 4 x + 5 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 5 = 0 : x = - 5

Solve

x^{2} - 4 x + 5 = 0 : x = 2 + i, x = 2 - i

The solutions are

x = 1, x = - 5, x = 2 + i, x = 2 - i

- 3 (2 x + 1)^{3} = - 192

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

x^{3} + x^{2} + 1

x^{4} + 8 x^{3} + 17 x^{2} + 8 x + 16 = 0

x^{4} + 12 x + 3 = 0

What are the solutions to the equation x^4=(4x-5)^2 ?
The solutions to the equation x^4=(4x-5)^2 are x=1,x=-5,x=2+i,x=2-i
Find the zeros of x^4=(4x-5)^2
The zeros of x^4=(4x-5)^2 are x=1,x=-5,x=2+i,x=2-i