<s>[INST] Finding values of $a$ with which two equations are equivalent; getting rid of radical sign Two equations are given:
$$x^2+(a^2-5a+6)x=0$$
$$x^2+2(a-3)x+a^2-7a+12=0$$
We need to find the values of $a$ that will render them equivalent. 
From the first equation,
$$x=-a^2+5a-6$$
From the second, 
$$x=\frac{-2a+6\pm2\sqrt{a-3}}{2}$$
If the two equations are equivalent, then x=x
$$-a^2+5a-6=\frac{-2a+6\pm2\sqrt{a-3}}{2}$$
That is, we must have two roots, considering the $\pm$ sign. From this, I've gotten as far as (for the plus sign case): 
$$2a^2-12a+18+2\sqrt{a-3}=0$$
For the minus case, there will be a minus before the term $a\sqrt{a-3}$. Basically these two equations will have the roots a=3 and a=4 respectively (from the answers section in my textbook). 
But how does one get to that? How to get rid of the square root sign? Or is there an alternative way to solve the whole shebang? 
 [/INST] Notice, we have $$x^2+(a^2-5a+6)x=0\tag 1$$
$$x^2+2(a-3)x+a^2-7a+12=0\tag 2$$
Now, since the coefficients of $x^2$ in both the equations is $1$ hence both the above equations will be equivalent to each other if their corresponding coefficients of $x$ & constant terms are equal hence, by comparison, we have the following cases $$\begin{cases} a^2-5a+6=2(a-3)\\ a^2-7a+12=0\end{cases}$$
or $$\begin{cases} a^2-7a+12=0\\ a^2-7a+12=0\end{cases}$$
or $$\begin{cases} (a-3)(a-4)=0\\ (a-3)(a-4)=0\end{cases}$$
Hence, we get $$a-3=0\iff a=3$$
or $$a-4=0\iff a=4$$ 
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