I had this question:

"Find the cubic equation whose roots are twice the roots of the equation $3x^3 - 2x^2 + 1 = 0$"

In my first attempt, I solved it through the use of simultaneous equations, where I let the cubic equation be: $x^3 + bx^2 + cx + d$. However, I've been told that there is a more efficient method for higher degree polynomials - and it has been labelled as "The Substitution method".

It involved something like "Let $y = x^2$" then form a new equation using that. Could someone solve this question and explain how to do it for me?

Any help would be greatly appreciated, thanks!

The solution with the substitution method worked like this:

Let $y=2x$, $x=y/2$ then it was substituted back into the given equation.

Can someone explain this?


Suppose you have a root $x$ of $$3x^3-2x^2+1=0$$

If $y=2x$, then $x=\cfrac y 2$ and $$3(\frac y 2)^3-2(\frac y 2)^2+1=0$$

Clear the fractions and you obtain an equation with integer coefficients satisfied by $y$.

The key to the substitution method is to find an expression for $x$ (a root of the original equation) and substitute it into the original equation. It depends on doing the same thing to each of the roots.

  • $\begingroup$ This cleared things up. Thanks! $\endgroup$ – missiledragon May 18 '13 at 8:07

We want the roots $y_i$ to be $2x_i$. So substitute $\frac{y}{2}$ everywhere for $x$, and simplify if you wish.

Afterwards, if you want the variable name to be $x$, replace $y$ by $x$.

  • $\begingroup$ Can you please clarify a bit more? $\endgroup$ – missiledragon May 18 '13 at 7:53

Let the roots of the given equation be $a_{1},a_{2},a_{3},a_{4}$.The new roots will be $2a_{1},2a_{2},2a_{3},2a_{4}$ Now use Vieta' formula. Vieta's Formulas

Sum of roots = $2(\sum a_{i})=\dfrac{4}{3}$ and product of roots = $8a_{1}a_{2}a_{3}=-8$

The equation will be : $$x^{3}-\frac{4}{3}x^{2}+\frac{8}{3}=0$$ or, $$3x^{3}-4x^{2}+8=0$$


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