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

The solutions to the equation (x+1)(x+2)(x+3)=x(x+4)(x+5) are x=-(3+sqrt(17))/2 ,x=(sqrt(17)-3)/2

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

The zeros of (x+1)(x+2)(x+3)=x(x+4)(x+5) are x=-(3+sqrt(17))/2 ,x=(sqrt(17)-3)/2

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(x + 1) (x + 2) (x + 3) = x (x + 4) (x + 5)

x = - \frac{3 + 17}{2}, x = \frac{17 - 3}{2}

Decimal Notation

x = - 3.56155 \dots, x = 0.56155 \dots

Solution steps

(x + 1) (x + 2) (x + 3) = x (x + 4) (x + 5)

Expand

(x + 1) (x + 2) (x + 3) : x^{3} + 6 x^{2} + 11 x + 6

Expand

x (x + 4) (x + 5) : x^{3} + 9 x^{2} + 20 x

x^{3} + 6 x^{2} + 11 x + 6 = x^{3} + 9 x^{2} + 20 x

Move

20 x

to the left side

x^{3} + 6 x^{2} - 9 x + 6 = x^{3} + 9 x^{2}

Move

9 x^{2}

to the left side

x^{3} - 3 x^{2} - 9 x + 6 = x^{3}

Move

x^{3}

to the left side

- 3 x^{2} - 9 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 3 ) \cdot 6}{2 ( - 3 )}

(- 9)^{2} - 4 (- 3) \cdot 6 = 317

x_{1, 2} = \frac{- ( - 9 ) \pm 3 17}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3 17}{2 ( - 3 )}, x_{2} = \frac{- ( - 9 ) - 3 17}{2 ( - 3 )}

x = \frac{- ( - 9 ) + 3 17}{2 ( - 3 )} : - \frac{3 + 17}{2}

x = \frac{- ( - 9 ) - 3 17}{2 ( - 3 )} : \frac{17 - 3}{2}

The solutions to the quadratic equation are:

x = - \frac{3 + 17}{2}, x = \frac{17 - 3}{2}

(t^{2} + 1) = 0

y, (y + 1)^{2} = 4 (x + 1)^{3}

x^{2} - 8 x + 22

x^{2} - 8 x + 20

x^{2} - 8 x + 21

What are the solutions to the equation (x+1)(x+2)(x+3)=x(x+4)(x+5) ?
The solutions to the equation (x+1)(x+2)(x+3)=x(x+4)(x+5) are x=-(3+sqrt(17))/2 ,x=(sqrt(17)-3)/2
Find the zeros of (x+1)(x+2)(x+3)=x(x+4)(x+5)
The zeros of (x+1)(x+2)(x+3)=x(x+4)(x+5) are x=-(3+sqrt(17))/2 ,x=(sqrt(17)-3)/2