Finding a quadratic equation given its two roots. Does this method exist?

Question

This is my first post in a math exchange, however not my first post in an "exchange" forum.

During one of my standardized testing preparation classes today, I was asked to find any quadratic equation given roots: $$ 1-2\sqrt{3} , 1+2\sqrt{3}. $$

This is a common problem in the particular standardized testing that I am preparing for. I came up with the following answer to the question in about 10 seconds: $$ (x-1)^2 = 12 $$ When solving: $$ \sqrt{(x-1)^2} = \sqrt{12} $$ $$ x-1 = \pm 2 \sqrt{3} $$ $$ x = 1 \pm 2 \sqrt{3} $$

All it took to solve the problem was to realize that $\pm 2\sqrt 3$ was the same as $\pm\sqrt{12}$. I like to think that it was a smart solution to the problem in question, and the reason why I bring up how much time it took me to do it is because once I thought of solving the problem in this way, it became very easy to think of solutions to similar problems. It is faster than factoring out.

Now, my question is: How could I relate this method of finding the quadratic equation in a problem where both roots are rational numbers? For example:

2 and -3

Would it even be useful to think of problems in this way? And lastly, does something similar to this already exist? All answers are appreciated as long as they are informative and/or helpful.

Thanks in advance.

Answer 1

The important thing about your method isn't that you have a radical, it's that you know the number directly between the two roots and the distance from each of the roots to that center number. You can solve your example problem in the same way. If the roots are $2$ and $-3$, you can rewrite these two roots as $-\frac{1}{2}\pm\frac{5}{2}$ ($-\frac{1}{2}$ is directly between $2$ and $-3$ and a distance of $\frac{5}{2}$ from both). Then we can solve the same way you did:

$\begin{align} x&=-\frac{1}{2}\pm\frac{5}{2} \\ x+\frac{1}{2}&=\pm\frac{5}{2} \\ \left(x+\frac{1}{2}\right)^2&=\left(\frac{5}{2}\right)^2 \\ \left(x+\frac{1}{2}\right)^2-\left(\frac{5}{2}\right)^2&=0 \end{align}$

So your quadratic equation is $f(x)=\left(x+\frac{1}{2}\right)^2-\left(\frac{5}{2}\right)^2$.

As other people are pointing out, the more obvious answer is $f(x)=(x-2)(x+3)$, but my method is how you can generalize the technique you provided. Note that both methods are equivalent, as $\left(x+\frac{1}{2}\right)^2-\left(\frac{5}{2}\right)^2$ can be rearranged to get $(x-2)(x+3)$.

Answer 2

If $x_1$ and $x_2$ are real/complex roots of a quadratic, then the quadratic is simply:

$$ a (x - x_1) (x-x_2) = 0 $$

with $a$ completely arbitrary except $0$.

Answer 3

Sum and product of roots are $ 2,-11\;$ respectively.

$$ \quad x^2-2x-11=0;\quad x= 1\pm 2 \sqrt 3 \text{ which roots come from }(x-1)^2=12; $$