What are the solutions to the equation (x^2+2x)^2-5(x^2+2x)+6=0 ?

The solutions to the equation (x^2+2x)^2-5(x^2+2x)+6=0 are x=1,x=-3,x=-1+sqrt(3),x=-1-sqrt(3)

Find the zeros of (x^2+2x)^2-5(x^2+2x)+6=0

The zeros of (x^2+2x)^2-5(x^2+2x)+6=0 are x=1,x=-3,x=-1+sqrt(3),x=-1-sqrt(3)

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(x^{2} + 2 x)^{2} - 5 (x^{2} + 2 x) + 6 = 0

x = 1, x = - 3, x = - 1 + 3, x = - 1 - 3

Decimal Notation

x = 1, x = - 3, x = 0.73205 \dots, x = - 2.73205 \dots

Solution steps

(x^{2} + 2 x)^{2} - 5 (x^{2} + 2 x) + 6 = 0

Expand

(x^{2} + 2 x)^{2} - 5 (x^{2} + 2 x) + 6 : x^{4} + 4 x^{3} - x^{2} - 10 x + 6

x^{4} + 4 x^{3} - x^{2} - 10 x + 6 = 0

Factor

x^{4} + 4 x^{3} - x^{2} - 10 x + 6 : (x - 1) (x + 3) (x^{2} + 2 x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x + 3 = 0 : x = - 3

Solve

x^{2} + 2 x - 2 = 0 : x = - 1 + 3, x = - 1 - 3

The solutions are

x = 1, x = - 3, x = - 1 + 3, x = - 1 - 3

a, a^{4} b + 2 c^{5} = d^{1} e

8 m^{3} - 104 m^{2} + 336 m = 0

5 x^{3} - 29 x + 30 = 0

- x^{3} + 3 x^{2} - 24 x + 1 = 0

x^{3} - 8 x^{2} + 21 x - 18 x = 0

What are the solutions to the equation (x^2+2x)^2-5(x^2+2x)+6=0 ?
The solutions to the equation (x^2+2x)^2-5(x^2+2x)+6=0 are x=1,x=-3,x=-1+sqrt(3),x=-1-sqrt(3)
Find the zeros of (x^2+2x)^2-5(x^2+2x)+6=0
The zeros of (x^2+2x)^2-5(x^2+2x)+6=0 are x=1,x=-3,x=-1+sqrt(3),x=-1-sqrt(3)