Why does $x + x = \dfrac{(-b)}{2a}$ not work fo both of the roots?

Question

So, the formula $x = \dfrac{(-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.

Answer 1

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.

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