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Bumbble Commhttps://math.techqa.club/user/bumbble-comm/detail
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I know that if $b^2-4ac>0$, then there are real solutions and so on.
But for any quadratic function, where is the discriminant present in the plot?
For example, $x^2-8x+7=f(x)$.
The square root of the discriminant is the distance between the zeros in case we have $a=1$. In your plot we have
$$7-1 = 6=\sqrt{36}=\sqrt{\Delta}.$$
Proof:
We have
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$$
so
$$x_1-x_2= \frac{-b+\sqrt{\Delta}}{2a} - \frac{-b-\sqrt{\Delta}}{2a} = \frac{2\sqrt{\Delta}}{2a}= \frac{\sqrt{\Delta}}{a}=\sqrt{\Delta}.$$
The square root of the discriminant is the distance between the zeros in case we have $a=1$. In your plot we have $$7-1 = 6=\sqrt{36}=\sqrt{\Delta}.$$
Proof:
We have $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$$ so $$x_1-x_2= \frac{-b+\sqrt{\Delta}}{2a} - \frac{-b-\sqrt{\Delta}}{2a} = \frac{2\sqrt{\Delta}}{2a}= \frac{\sqrt{\Delta}}{a}=\sqrt{\Delta}.$$