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what's the proof for the discriminant $b^2-4ac$?

e.g. Proove that if $b^2 > 4ac$ then x has 2 real and distinct roots.

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I don't understand what you mean by the proof for the discriminant. I assume u want to show that discriminant is $b^2-4ac$

$ax^2+bx+c$ and equate it to $0$

$ax^2+bx+c=0$

$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0$

Completeting the square, we get,

$(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}=0$

$(x+\frac{b}{2a})^2}=\frac{b^2}{4a^2}-\frac{c}{a}$

$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}$

$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

the $b^2-4ac$ is defined to be the discriminant.

the second part of the questions is obvious from here

$If b^2>4ac then \sqrt{b^2-4ac}>0$

$Making the roots of the quadratic x_1,x_2 real$

The distinct bit also can be seen clearly from here. I will leave it to the author of the question to write up the proper solution.

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Well there isn't a "proper" solution really. It's what you said, essentially ##### Share on other sites that quadratic form proof above show the result of one's solution. but the "discriminant" which is what under the sqrt show ahead what kind of root you will have. if a neg discriminant, then youll get a complex solution. if its zero, then youll have two solution thats equal. if its pos, then youll have two unequal solutions.

but then you say how do they know to use the equation under the sqrt to check for the discriminant. shoot(who knows), trial and error?

thats my 2 cents.

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You use the equation under the square root because the only way that the whole quadratic formula can be complex is if you have a negative under a square root. This is the only square root I see so it makes sense to look there. Also since it is +/- the only way that you can have one solution multiplicity two (two solutions that are the same number) is if the discriminant is equal to zero. Also, to clarify the quad form gives you both solutions. That is where the +/- comes in.

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