# If $3x^2-6x+p=0$ has roots $\alpha$ and $\beta$, then find a quadratic with roots $(\alpha+\beta)/\alpha$ and $(\alpha+\beta)/\beta$

This one comes from an IGCSE Pure Math past paper:

The equation $3x^2-6x+p=0$ has roots $\alpha$ and $\beta$. Without solving the equation, form a quadratic equation with roots $\frac{\alpha+\beta}{\alpha}$ and $\frac{\alpha+\beta}{\beta}$.

I solved it this way:

Given, $a=3, b=-6, c=p$

Sum of roots ($\alpha+\beta$) = $-b/a$ = $2$

Product of roots ($\alpha\beta$) = $c/a$ = $\frac{p}{3}$

When roots are $\frac{\alpha+\beta}{\alpha}$ and $\frac{\alpha+\beta}{\beta}$,

$S=(\frac{\alpha+\beta}{\alpha})+(\frac{\alpha+\beta}{\beta}$), which can be readily simplified to $\frac{(\alpha+\beta)^2}{\alpha\beta}$. Keep in mind you need to keep it in sums and products of the roots as that are the only stuff you know and can substitute in to get a unknown-free result.

And so $S=\frac{(2)^2}{(\frac{p}{3})}=\frac{12}{p}$.

The product of the new roots is also $\frac{12}{p}$, since $P=(\frac{\alpha+\beta}{\alpha})\times(\frac{\alpha+\beta}{\beta})=\frac{(\alpha+\beta)^2}{\alpha\beta}$.

Using the principle that, if $S=\text{sum of roots}$ and $P=\text{product of roots}$, the equation will be $x^2-x(S)+P=0$:

I end up with $\boxed{px^2-12x+12=0}$, after multiplying out the equation with a common LCM to rid it of fractions.

The question is way ancient and I can't find the mark scheme online. I need confirmation on whether or not this is the right approach and the final answer is correct. I'm only an 8th grader, so there are defintely things in the IGCSE syllabus I don't know much about.

Feel free to edit the tags; I don't have much of an idea on what tags could be suitable. So I just put somewhat random tags.

Yes it is a correct way indeed

$$\left(x-\frac{\alpha+\beta}{\alpha}\right)\left(x-\frac{\alpha+\beta}{\beta}\right)=x^2+\left(\frac{(\alpha+\beta)^2}{\alpha\beta}\right)x+\frac{(\alpha+\beta)^2}{\alpha\beta}$$

• Ooh...this is even more brilliant! You did it in just one line. Being with you guys makes me feel stupid. That's part of the reason I don't venture out to MSE, unless I really need to. :-P – Soha Farhin Pine Apr 2 '18 at 8:18
• @SohaFarhinPine Your work is very good, the nice thing here on MSE is that there is always something to learn for everybody! Bye – user Apr 2 '18 at 8:24

You are correct.

Note that $\displaystyle \frac{\alpha+\beta}{\alpha}=\frac{2}{\alpha}$ and $\displaystyle \frac{\alpha+\beta}{\beta}=\frac{2}{\beta}$.

So $\displaystyle S=\frac{2}{\alpha}+\frac{2}{\beta}=\frac{2(\alpha+\beta)}{\alpha\beta}=\frac{2(2)}{p/3}=\frac{12}{p}$ and $\displaystyle P=\frac{2}{\alpha}\times\frac{2}{\beta}=\frac{4}{\alpha\beta}=\frac{4}{p/3}=\frac{12}{p}$.

Other Method:

As $\alpha$ is a root of $3x^2-6x+p=0$,

\begin{align*} 12\alpha^2-24\alpha+4p&=0\\ p\left(\frac{2}{\alpha}\right)^2-12\left(\frac{2}{\alpha}\right)+12&=0 \end{align*}

Similarly, $\displaystyle p\left(\frac{2}{\beta}\right)^2-12\left(\frac{2}{\beta}\right)+12=0$

So, $\displaystyle \frac{2}{\alpha}$ and $\displaystyle \frac{2}{\alpha}$ are the roots of $px^2-12x+12=0$.

Yes, your approach is correct :)