# Why does x + x = (-b)/2a not work fo both of the roots?

So, the formula x = (-b)/2a is used to find the x value of the axis of symmetry in a quadratic equation. and since the x value will always be exactly half of the 2 roots and it seems to work I can just add x to itself after finding it and find one of the roots of the quadratic equation but it doesn't work to get the other root I simply thought I will be able to subtract x from itself but it will give me 0 unless it's negative. Sorry if this is a stupid question but I Love math and like to experiment with it.

• The sentence "the $x$ value will always be exactly half of the 2 roots" doesn't make sense. If $r_1$ and $r_2$ are the roots of your quadratic equation then the $x$-value of the axis of symmetry equals $(r_1+r_2)/2$, that is, the $x$-value of the midpoint of the segment between $r_1$ and $r_2$ on the $x$-axis. – Must Apr 20 '18 at 2:14
• @Must Alright that makes sensei was stupid thanks man and nice explanation – k4b00m Apr 20 '18 at 2:16
• The roots sum to $-b/a.$ – Chris Leary Apr 20 '18 at 2:17
• @Must So is there any way to figure out the root of the equation using the relationship between the axis of symmetry and the equation you showed above. I did set up this equation but still doesn't work : (r1+r2)/2 = x so: r1 = 2x-r2 and then using substitution I tried it but didn't work and in this whole scenario I am assuming know the "x" or the x coordinate of the axis of symmetry. – k4b00m Apr 20 '18 at 2:30

## 1 Answer

The figure shows the relative positions of the roots ($\alpha$ the smaller; and $\beta$ the larger) and the axis of symmetry. If we express all the unknown quantities in terms of $\alpha$ and $\beta$, then the quantity to be added to (or subtracted from) $\dfrac {-b}{2a}$ is s = … = $\dfrac {\beta - \alpha}{2}$, which is the constant you are looking.

In fact, after some root sum/product calculation, $s = … = \sqrt {\dfrac {b^2 – 4ac}{2a}}$. Then, $\beta = \dfrac {-b}{2a} + \sqrt {\dfrac {b^2 – 4ac}{2a}}$, which is just back to square 1, the original quadratic formula.

Added: As seen in the figure, the constant you are looking for and supposed to be added or subtracted is s = $\dfrac {\beta - \alpha}{2}$.

Note that $(\beta - \alpha)^2 = … = (\beta + \alpha)^2 – 4\beta \alpha = (\dfrac {-b}{a})^2 – 4(\dfrac {c}{a}) = … = \dfrac {b^2 – 4ac}{a^2}$.

• Ok but why are we using sqrt(b^2-4ac) and how is this output added or subtracted from the x component gives us the roots – k4b00m Apr 20 '18 at 10:36
• @k4b00m Pls see added. – Mick Apr 21 '18 at 13:36