Prove this without completing the square

[eqn](x_1+x_2)^2-4x_1x_2=(x_1-x_2)^2[/eqn]

The result follows trivially from this identity.

We all have our days, man.

im not the bloke who wrote that but:

[math]a^2 + 2ab + b^2 = (a + b)^2[/math]

based wolfram poster

The fact to the matter is that these equations are not numerically stable, please use different ones if you actually numerically calculate your results.
2/10, see me after class

show it then, if it's so trivial

[eqn](x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2[/eqn]
Using the identity, you can write both the sum and difference of the roots in terms of the coefficients. You just need to solve those two linear equations to get the individual roots. Trivial.

you are still completing the square here bro...

>without completing the square
>brainlet used completing the squared technique

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