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

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[MarkFL]

We are given the cubic polynomial:

\(\displaystyle x^3+3x^2+2=0\)

And we are instructed to use the substitution:

\(\displaystyle x=\frac{1}{\sqrt{u}}\)

And so the cubic becomes:

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

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

\(\displaystyle 1+3\sqrt{u}+2u\sqrt{u}=0\)

Next, let's arrange as:

\(\displaystyle \sqrt{u}=-\frac{1}{3+2u}\)

Square both sides:

\(\displaystyle u=\frac{1}{9+12u+4u^2}\)

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:

\(\displaystyle 9u+12u^2+4u^3-1=0\)

Arrange in standard form:

\(\displaystyle 4u^3+12u^2+9u-1=0\)

 

[Harry2]

Thank you! I was constantly getting the wrong answer.

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

 

[MarkFL]

Thank you! I was constantly getting the wrong answer.

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

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