Finding the value of a parameter, when equations have a common root

Question

How can I find the value of the real parameter $m$, if

$2x^2-3x+1=0$ and $3x^2+m(x+2)+1=0$

have a common root?

I opened the parentheses, got rid of the $x^2$ , but I get a big fraction as the value of the common root. Where did I go wrong?

Answer 1

Let $a$ be the common root

$$\implies 2a^2-3a+1=0\ \ \ \ (1)\implies a=?$$

Set the value of $a$ in $$3a^2+m(a+2)+1=0\ \ \ \ (2)$$ to find $m$

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Alternatively, solve $(1),(2)$ for $a,a^2$ and use $a^2=(a)^2$

Answer 2

The roots of the first equation $$2x^2-3x+1= 0$$ are $1$ and $\frac{1}{2}$.

Rewrite the second equation as $$3x^2+mx+2m+1=0$$ and use the quadratic formula to find its roots as follows.

$$x=\frac{-m\pm \sqrt{m^2-(4)(3)(2m+1)}}{6}$$ $$x=\frac{-m\pm \sqrt{m^2-24m-12}}{6}$$

Then solve $$\frac{-m\pm \sqrt{m^2-24m-12}}{6}=\frac{1}{2}$$ $$\frac{-m\pm \sqrt{m^2-24m-12}}{6}=1$$ for $m$.​