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

The solutions to the equation (x^2+x)^2=5x^2+5x-6 are x=1,x=-2,x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2

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

The zeros of (x^2+x)^2=5x^2+5x-6 are x=1,x=-2,x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2

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

x = 1, x = - 2, x = \frac{- 1 + 13}{2}, x = \frac{- 1 - 13}{2}

Decimal Notation

x = 1, x = - 2, x = 1.30277 \dots, x = - 2.30277 \dots

Solution steps

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

Expand

(x^{2} + x)^{2} : x^{4} + 2 x^{3} + x^{2}

x^{4} + 2 x^{3} + x^{2} = 5 x^{2} + 5 x - 6

Move

6

to the left side

x^{4} + 2 x^{3} + x^{2} + 6 = 5 x^{2} + 5 x

Move

5 x

to the left side

x^{4} + 2 x^{3} + x^{2} + 6 - 5 x = 5 x^{2}

Move

5 x^{2}

to the left side

x^{4} + 2 x^{3} - 4 x^{2} - 5 x + 6 = 0

Factor

x^{4} + 2 x^{3} - 4 x^{2} - 5 x + 6 : (x - 1) (x + 2) (x^{2} + x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

x^{2} + x - 3 = 0 : x = \frac{- 1 + 13}{2}, x = \frac{- 1 - 13}{2}

The solutions are

x = 1, x = - 2, x = \frac{- 1 + 13}{2}, x = \frac{- 1 - 13}{2}

lo g_{2} (256) = \frac{x ^{3}}{8}

4 x^{4} - 10 x^{3} - 1 x^{2} = 0

3 x^{4} - 8 x^{3} - 16 x^{2} + 24 x + 32 = 0

6 x^{3} - 7 x^{2} - 6 x = - 7

z^{3} - 4 z^{2} + 9 z - 10 = 0

What are the solutions to the equation (x^2+x)^2=5x^2+5x-6 ?
The solutions to the equation (x^2+x)^2=5x^2+5x-6 are x=1,x=-2,x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2
Find the zeros of (x^2+x)^2=5x^2+5x-6
The zeros of (x^2+x)^2=5x^2+5x-6 are x=1,x=-2,x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2