# Solve for x:  6x = 2x^3 + 11x^2

sciencesolve | Certified Educator

You need to factor out `x^2 ` to the right side, such that:

`6x = 2x^3 + 11x^2 => 6x = x^2(2x + 11)`

You need to subtract `x^2(2x + 11)` both sides, such that:

`6x - x^2(2x + 11) = 0 `

Factoring out x, yields:

`x(6 - x(2x + 11)) = 0`

Using the zero product property yields:

`x(6 - x(2x + 11)) = 0 => {(x = 0),(6 - x(2x + 11) = 0):}`

`6 - x(2x + 11) = 0 => 6 - 2x^2 - 11x = 0`

Rearranging the terms yields:

`-2x^2 - 11x + 6 = 0`

Factoring out -1 yields:

`-(2x^2 + 11x - 6) = 0 => 2x^2 + 11x - 6 = 0`

`2x^2 + 11x = 6`

Completing the square to the left side yields:

`2x^2 + 11x + 121/8 = 6 + 121/8 `

`(sqrt2*x + 11/(2sqrt 2))^2 = 169/8 => sqrt2*x + 11/(2sqrt 2) = +-sqrt(169/8) => sqrt2*x = -11/(2sqrt 2) +- 13/(2sqrt 2)`

`x_(1,2) = (-11 +- 13)/4 => x_1 = 1/2 ; x_2 = -6`

Hence, evaluating the solutions to the given cubic equation yields `x_1 = 0, x_2 = 1/2 ; x_3 = -6.`

mariloucortez | Certified Educator

Since the degree of equation is in the third degree (because 3 is the greatest exponent), you'll be expecting 3 solutions for this problem.

Rewrite the equation in general form, such that all the terms are on one side and the other is just 0.

`2x^3 + 11x^2 -6x = 0`

Notice that 6x became -6x. It is because it is moved to the other side to be with the other terms.

Factor out x, since it is present in all the terms.

`x(2x^2 + 11x - 6) = 0`

You can still factor out the terms inside () by quadratic formula since it is in the form `ax^2+bx+c` :

`x = (-b +-sqrt(b^2 -4ac))/(2a)`

`x = (-11+- sqrt(11^2 - 4(2)(-6)))/(2*2)`

` ` `x=(-11+-sqrt(169))/4`

`x = (-11+-13)/4`

`x=(-11+13)/4`

`x=2/4 = 1/2`

`x=(-11 - 13)/4`

`x=-24/4 =- 6`

Therefore the solutions are -6, 0 and `1/2.`

justaguide | Certified Educator

The equation 6x = 2x^3 + 11x^2 has to be solved for x. As this is a cubic equation there are 3 solutions for x.

6x = 2x^3 + 11x^2

=> 2x^3 + 11x^2 - 6x = 0

=> x(2x^2 + 11x - 6) = 0

=> x(2x^2 + 12x - x - 6) = 0

=> x(2x(x + 6) - 1(x + 6)) = 0

=> x(2x - 1)(x + 6) = 0

`2x - 1 = 0 => x = 1/2`

`x + 6 = 0 => x = -6`

And the third root is x = 0

The solution of the given equation is {-6, 0, 1/2}

pramodpandey | Student

We have

`6x=2x^3+11x^2`   ,let us write it as

`2x^3+11x^2-6x=0`  ,factor out x from it

`x(2x^2+11x-6)=0`  ,product of two number is zero.

`x=0 `   or `2x^2+11x-6=0`

`2x^2+11x-6=0`

`2x^2+12x-x-6=0`

`2x(x+6)-1(x+6)=0`

`(x+6)(2x-1)=0`

`x+6=0 ,or `

`2x-1=0, `

`x=-6 ,and 2x=1`

`x=1/2`

` x=0 ,x=1/2, x=-6`

`Ans.`