Quadratic Formula – When and How To Solve Quadratic Equations with the Quadratic Formula!

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[joubess]

Discriminant, tells if your solutions are real or imaginary. If your discriminant is negative, you take the square root of a negative number, so you get a complex value with an i in the solutions.

[Stylux-zp]

side notes : sqrt means square root, this ^ symbol means raise to the power of, +/- means plus or minus
x^2 - 4x = 8
method used: Completing the square
Take the coefficient of the second term divide it by 2 and square it (-4/2)^2 = (16/4) = 4 substitute this in equation below
x^2 - 4x + ... = 8 + ...
x^2 - 4x + 4 = 8 + 4
x^2 - 4x + 4 = 12
Take the coefficient of the second term divide it by 2 and substitute it below -4/2 = -2
(x + ..)^2 = 12
(x - 2)^2 = 12
sqrt((x - 2)^2 ) =+/- sqrt(12)
x - 2 = +/-sqrt(12)
x - 2 = +/- sqrt( 4 * 3)
x - 2 = +/- 2sqrt(3)
x = +/- 2sqrt(3) + 2

[ralphmelvin]

This all seems crazy to me. All this work Algebra 2 progressive algebra is crazy to me.. I appreciate it when you say keep trying if you don't understand algebra just keep trying you're so right there that's what I have to do.. the quadratic formula confuses me.. but you are a very good teacher you take it step by step people criticize you for being too slow but I like the way you teach you teach very well

[russelllomando]

wow got it fun with negatives thanks...

[tomtke]

{-B +/- sqrt(B^2 -4AC)} ÷ 2A
x^2-4x-8 = 0
A=1
B=-4
C=-8


{-B+/-sqrt(B^2-4AC)}÷2A
=>
{4+/-sqrt(16-4•1•(-8))}÷2•1
=
{4+/-sqrt(16+32)}÷2
{4+/-sqrt(48)}÷2
{4+/-4sqrt(3)}÷2
2+/-2sqrt(3)
sol.1 2+2sqrt(3)
sol.2 2-2sqrt(3)
i.e.

=>

VERIFY


x^2

+(2+2sqrt(3))(2-2sqrt(3))
=
x^2
-4x
+{4-4sqrt(3)+4sqrt(3)-12}
=
x^2
-4x
-8
=0
=>
x^2-4x-8=0
or
x^2-4x=❤8✔️

[argonwheatbelly]

2(1±√3) in my head in about a minute. Too slow, I know. 😞