How to solve the equation $x^2 - 6x + 25 = 0$

Question

There is one thing that I don't understand.

How is it possible that $-6x$ can be into "$9$".

Can you describe it to me by calculation?

Thanks!

Answer 1

Notice, in general roots of $ax^2+bx+c=0$ are calculated as follows $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Now, we have $$x^2-6x+25=0$$ Using square formula, we get $$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(25)}}{2(1)}$$ $$=\frac{6\pm\sqrt{-64}}{2}$$ $$=\frac{6\pm8i}{2}=3\pm4i$$

Hence, the roots are complex

Answer 2

Since $6$ is even they use a modified version of the quadratic formula: $$ax^2 + 2b'x +c=0$$ $$\Delta ' = b'^2 - a*c$$ $$x=\frac {-b' \pm \sqrt{\Delta'}}{a}$$

The $2$ is simplified from the original formula.

Answer 3

Consider $x^2-6x+25=(x-3)^2+16$. Here $16=25-9$

This is a general phenomenon. $x^2-2mx+n=(x-m)^2+n-m^2$. The $9$ comes from the fact that in your case $2m=6$ and $m^2=9$.

The point is that $(x-3)^2=x^2-6x+9$