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While I was applying the quadratic formula, it just suddenly dawned to me that the quadratic formula was basically finding the vertex and adding or subtracting (the plusminus) half the distance between the roots. $\frac{-b}{2a}$ represents the vertex and the $\pm \frac{\sqrt{\triangle}}{2a}$ represents half the distance between the roots. I understand why $-\frac{b}{2a}$ is the xcoor of the vertex, it intuitively makes sense. But why does the $\frac{\sqrt{\triangle}}{2a}$ equal half the distance between the roots? i know how to prove it but I dont why it intuitively works. Edit: I understand why the $\frac{\sqrt{\triangle}}{2a}$ is half the distance from the vertex but why is the discriminant B^2 - 4ac. I know thats what happens when you complete the square but when you use algebra you often lose the intuitive sense of what's happening.

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    $\begingroup$ One definition of discriminant would, for a quadratic, be $\Delta=a^2(\alpha-\beta)^2$ where $\alpha,\beta$ are the roots and $a$ is the leading coefficient. As such. the distance $|\alpha-\beta|$ can be obtained as $\sqrt{\Delta}/a$ (when $\alpha,\beta$ are real numbers anyway). $\endgroup$ – anon Dec 17 '17 at 23:07
  • $\begingroup$ @anon : Very good comment that should become an answer. Especially clear for people who are acquainted with the two ways one can write disc(f)=res(f(x),f'(x)). See (www2.math.uu.se/~svante/papers/sjN5.pdf). $\endgroup$ – Jean Marie Dec 17 '17 at 23:34
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It may be helpful to think about this in terms of symmetries and transformations of graphs. Let $f(x) = x^2$. The graph of $f$ is our primitive parabola. Observer that $f$ is an even function, i.e. $$ f(-x) = x^2 = f(x) $$ for all real numbers $x$. This means that the graph of $f$ is symmetric about the $y$-axis. The vertex of this graph is at the origin.

Now consider shifting this graph down $k$ units. That is, consider the graph of the function $x \mapsto x^2 - k$, where $k \ge 0$. As this is just a basic transformation, it doesn't affect the underlying symmetry of the graph. Indeed, this new function is still symmetric across the $y$-axis. But observe that $$ x^2 - k = (x-\sqrt{k})(x+\sqrt{k}), $$ hence the roots of this function are $\pm\sqrt{k}$. Also observe that, thanks to the underlying symmetry, the two roots are equidistant from the vertex, which implies that the midpoint of the segment joining the roots is on the axis of symmetry.

Next, shift this graph horizontally. For some real number $h$, consider the function $x \mapsto (x-h)^2 + k$. This moves the vertex to the point $(h,k)$, but preserves the basic symmetry of the graph—the axis of symmetry is now the line $x = h$, and the roots are $h\pm k$. As before, the midpoint of the roots is on the axis of symmetry.

Finally, a completely general quadratic function is given by $$ A(x-h)^2 + k.$$ This is obtained by scaling the previous function by a factor of $A$ (and absorbing a factor of $A$ into the vertical shift). We still haven't done anything that breaks the original symmetry, hence the roots of this (which are given by $$ x = h \pm \sqrt{-\frac{k}{A}}, $$ from which we can obtain the quadratic formula, if we like) are again equidistant from the vertex, and have midpoint on the axis of symmetry.

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That is because, when you complete the square, $\;\Bigl(x+\dfrac b{2a}\Bigr)^2$ is the square of the $x$-distance from the point with abscissa $x$ to the point with abscissa $-\dfrac b{2a}$, i.e. to the vertex of the parabola.

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  • $\begingroup$ I've seem to have worded my question badly. I reupdated my question $\endgroup$ – Joseph Lee Dec 18 '17 at 0:41
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When you complete the square you can rewrite the equation $ax^2+bx+c=0$ (with $a>0$) as $$ \left(x+\frac{b}{2a}\right)^{\!2}=\frac{b^2-4ac}{4a^2} $$ If you set $X=x+\frac{b}{2a}$ then the equation gets the form $$ X^2=\frac{b^2-4ac}{4a^2} $$ For an equation $X^2=c$, the difference of the roots (larger minus smaller) is obviously $2\sqrt{c}$; a shift of the unknown doesn't change the difference of the roots. So the difference of the roots is $$ 2\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\sqrt{b^2-4ac}}{a} $$ If you only assume $a\ne0$, the difference should be written $$ \frac{\sqrt{b^2-4ac}}{|a|} $$

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  • $\begingroup$ I've seem to have worded my question badly. I reupdated my question $\endgroup$ – Joseph Lee Dec 18 '17 at 0:41

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