What are the solutions to the equation (x^2-4x+5)^2-(x-1)(x-3)=4 ?

The solutions to the equation (x^2-4x+5)^2-(x-1)(x-3)=4 are x=1,x=3,x=2+sqrt(2)i,x=2-sqrt(2)i

Find the zeros of (x^2-4x+5)^2-(x-1)(x-3)=4

The zeros of (x^2-4x+5)^2-(x-1)(x-3)=4 are x=1,x=3,x=2+sqrt(2)i,x=2-sqrt(2)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^{2} - 4 x + 5)^{2} - (x - 1) (x - 3) = 4

x = 1, x = 3, x = 2 + 2 i, x = 2 - 2 i

Solution steps

(x^{2} - 4 x + 5)^{2} - (x - 1) (x - 3) = 4

Expand

(x^{2} - 4 x + 5)^{2} - (x - 1) (x - 3) : x^{4} - 8 x^{3} + 25 x^{2} - 36 x + 22

x^{4} - 8 x^{3} + 25 x^{2} - 36 x + 22 = 4

Move

4

to the left side

x^{4} - 8 x^{3} + 25 x^{2} - 36 x + 18 = 0

Factor

x^{4} - 8 x^{3} + 25 x^{2} - 36 x + 18 : (x - 1) (x - 3) (x^{2} - 4 x + 6)

(x - 1) (x - 3) (x^{2} - 4 x + 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 3 = 0 or x^{2} - 4 x + 6 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 3 = 0 : x = 3

Solve

x^{2} - 4 x + 6 = 0 : x = 2 + 2 i, x = 2 - 2 i

The solutions are

x = 1, x = 3, x = 2 + 2 i, x = 2 - 2 i

2 x^{4} - 29 x^{2} - 48 = 0

- x^{3} - 2 x^{2} + 8 x = 0

x, y = 8 x^{3}

x^{4} - 50 x^{2} + 625 = 0

x^{3} - 3 x^{2} + 5 = 0

What are the solutions to the equation (x^2-4x+5)^2-(x-1)(x-3)=4 ?
The solutions to the equation (x^2-4x+5)^2-(x-1)(x-3)=4 are x=1,x=3,x=2+sqrt(2)i,x=2-sqrt(2)i
Find the zeros of (x^2-4x+5)^2-(x-1)(x-3)=4
The zeros of (x^2-4x+5)^2-(x-1)(x-3)=4 are x=1,x=3,x=2+sqrt(2)i,x=2-sqrt(2)i