MHB Using Substitution to Solve Cubic Equations

[Harry2]

Hi,
I don't understand how to get to the answer of the question. The cubic equation x^3 + 3(x^2) +2 =O.by using substitution X=1/(u^0.5) get 4u^3+12u^2+9u-1 =0.

I can't see where the 12 comes in
It's question 10i

View attachment 5172

View attachment 5173

 

[MarkFL]

We are given the cubic polynomial:

$$x^3+3x^2+2=0$$

And we are instructed to use the substitution:

$$x=\frac{1}{\sqrt{u}}$$

And so the cubic becomes:

$$\left(\frac{1}{\sqrt{u}}\right)^3+3\left(\frac{1}{\sqrt{u}}\right)^2+2=0$$

If we multiply through by $u\sqrt{u}$, we get:

$$1+3\sqrt{u}+2u\sqrt{u}=0$$

Next, let's arrange as:

$$\sqrt{u}=-\frac{1}{3+2u}$$

Square both sides:

$$u=\frac{1}{9+12u+4u^2}$$

Do you see how the 12 came from the squaring of the binomial $3+2u$?

Multiply through by $9+12u+4u^2$ and then subtract through by 1, to obtain:

$$9u+12u^2+4u^3-1=0$$

Arrange in standard form:

$$4u^3+12u^2+9u-1=0$$

 

[Harry2]

Thank you! I was constantly getting the wrong answer.

- - - Updated - - -

Why though do you factorise the u^.5. Can't you just square it?

 

[MarkFL]

Thank you! I was constantly getting the wrong answer.

- - - Updated - - -

Why though do you factorise the u^.5. Can't you just square it?

Isolating $\sqrt{u}$ makes the squaring process simpler. :)