# Given that -0.66 is the solution of the equation 6x^3=10x(2x+1), determine the other solutions?

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### 2 Answers

Q:Given the equation 6x^3=10x(2x+1), determine the solutions (edited).

Solution:

The given equation is 6x^3=10x(2x+1).

Therefore 6x^3-20x-10x = 0.

= > 2x(3x^2-10x-5) = 0.

Or x= 0, Or 3x^2-10x-5 = 0

We know that ax^2+bx+c = 0 has the factors x1 = {-b+sqrt(b^2-4ac)}/2a or x2 = { -b- sqrt(b^2-4ac)}/2a.

x^2-10x-1 = 0 has the roots:

x1 = {-(-10)+sqrt(10^2-4*3*-5)}/2*3 = (10 + sqrt(160)}6 = 3.7749.

Or x1 = (5+sqrt2sqr10)/3

x2 = (5-2sqrt10)/3

**Therefore the solutions of 6x^3= 10x(2x+1) are:**

** x= 0, **

**x1 = (5+2sqrt10)/3 **

**x2 = (5-2sqrt10)/3.**

Since the equation is of 3rd order, that means that it has 3 roots. If one root is known, the other 2 roots are x,y.

We'll write the given root as a fraction:

-0.66 = -2/3

We'll divide the equation by 2 and we'll get:

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

We'll remove the brackets:

3x^3 = 10x^2 + 5x

We'll move all terms to the left side:

3x^3 - 10x^2 - 5x = 0

-24/27 - 40/9 + 10/3=0

-24-120+90 = -54

As we can notice, -0.66 is not the root of the equation.

We'll find the real roots:

We'll factorize by x:

3x^3 - 10x^2 - 5x = 0

x(3x^2 - 10x - 5) = 0

The first root is x1 = 0

Now, we'll apply quadratic formula to determine the other roots:

3x^2 - 10x - 5 = 0

x2 = [10+sqrt(100 + 60)]/6

x2 = (10+4sqrt10)/6

x2 = (5+2sqrt10)/3

x3 = (5-2sqrt10)/3

**The roots of the equation are: {(5-2sqrt10)/3 ; 0 ; (5+2sqrt10)/3}.**