# 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?

You omitted the second solution to the first equation, $x=0$. For the two equations to be equivalent, $x=0$ must also satisfy the second equation. That yields $a^2-7a+12=0$, which factors nicely into $(a-3)(a-4)=0$. Substituting $a=3$ and $a=4$ into the equations shows that they are equivalent in both cases.

In this particular case, the person who created the problem provided a clue by using $x=0$ as one of the solutions, and I suspect they did this because you're not expected to know the more general theory, but in case you're interested: A quadratic equation is determined by its roots up to a scale factor, so if you scale both equations such that the coefficient of the quadratic term is $1$ (in this case they're already thus scaled), you can simply equate the coefficients of the linear and constant terms, since they have to be equal if the roots are equal. In this particular case this approach was suggested by the root $x=0$, but it's valid in the general case, too.

• Than you, Joriki! Very interesting! I'll try to google this up. Aug 18, 2015 at 9:45

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 $$a-4=0\iff a=4$$

• You may be assuming more theory here than the OP might have available. The clue structure of the problem seems to suggest that it's not taken as known that the coefficients are determined by the roots up to a scale factor. (Also you didn't mention the scale factor.) Aug 18, 2015 at 6:21
• (Now you did. ) Aug 18, 2015 at 6:32