What are the solutions to the equation (x^2+4x)^2-17(x^2+4x)=60 ?

The solutions to the equation (x^2+4x)^2-17(x^2+4x)=60 are x=-1,x=-3,x=2(sqrt(6)-1),x=-2(1+sqrt(6))

Find the zeros of (x^2+4x)^2-17(x^2+4x)=60

The zeros of (x^2+4x)^2-17(x^2+4x)=60 are x=-1,x=-3,x=2(sqrt(6)-1),x=-2(1+sqrt(6))

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(x^{2} + 4 x)^{2} - 17 (x^{2} + 4 x) = 60

x = - 1, x = - 3, x = 2 (6 - 1), x = - 2 (1 + 6)

Decimal Notation

x = - 1, x = - 3, x = 2.89897 \dots, x = - 6.89897 \dots

Solution steps

(x^{2} + 4 x)^{2} - 17 (x^{2} + 4 x) = 60

Expand

(x^{2} + 4 x)^{2} - 17 (x^{2} + 4 x) : x^{4} + 8 x^{3} - x^{2} - 68 x

x^{4} + 8 x^{3} - x^{2} - 68 x = 60

Move

60

to the left side

x^{4} + 8 x^{3} - x^{2} - 68 x - 60 = 0

Factor

x^{4} + 8 x^{3} - x^{2} - 68 x - 60 : (x + 1) (x + 3) (x^{2} + 4 x - 20)

(x + 1) (x + 3) (x^{2} + 4 x - 20) = 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 - 20 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x + 3 = 0 : x = - 3

Solve

x^{2} + 4 x - 20 = 0 : x = 2 (6 - 1), x = - 2 (1 + 6)

The solutions are

x = - 1, x = - 3, x = 2 (6 - 1), x = - 2 (1 + 6)

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

x, x^{3} - 6 x^{2} + 6 x + 5 = 0

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

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

x^{4} - 144 = 0

What are the solutions to the equation (x^2+4x)^2-17(x^2+4x)=60 ?
The solutions to the equation (x^2+4x)^2-17(x^2+4x)=60 are x=-1,x=-3,x=2(sqrt(6)-1),x=-2(1+sqrt(6))
Find the zeros of (x^2+4x)^2-17(x^2+4x)=60
The zeros of (x^2+4x)^2-17(x^2+4x)=60 are x=-1,x=-3,x=2(sqrt(6)-1),x=-2(1+sqrt(6))