Find the discriminant of ( x – 2 ) ( x + 4 ) = 2x?

(x-2)(x+4)=2x

x2-2x+4x-8=2x

X2+2x-8=2x
x2+2x-2x=8
x2=8
x=root of 8

use b^2- 4 a c
first expand
x^2 +4x-2x-8=2x , x^2+2x -2x -8 , x^2 +0x -8
=x^2 + 0x -8
=(0)^2-4(1)(-8) x is 1(x)
=0+32
=32

(x - 2)(x + 4) = 2x
x*x - 2*x + x*4 - 2*4 = 2x
x^2 - 2x + 4x - 8 - 2x = 0
x^2 + 2x - 2x - 8 = 0
x^2 + 0x - 8 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 0
c = -8

x = [0 ±√(0^2 + 32)]/2
x = [±√32]/2

∴ the discriminant is 32.

x ² + 2x - 8 = 2x
x² - 8 = 0
x ² = 8
x = ± 2√2

x^2-2x+4x+16=2x
x^2+16=0
b^2-4ac=discriminant
where b=0 a=1 c=16
therefore, D=0^2-4(1*16)
= -4*16
= -32

x(square) + 2x -8
=(2)(Square)-4(1)(-8)
=4-12
=8

Foil it out:
x^2 + 2x - 8 = 2x

Bring the 2x to the left side:
x^2 - 8 = 0 (the 2x cancels out on both sides)

Now use: b^2 - 4ac
a = 1
b = 0
c = -8

(0)^2 - 4(1)(-8)
0 + 32

Discriminant is 32

Hope this helps you
:o)
d

32 you definetly just got your hw done in like 3 seconds

x^2 + 2x - 8 = 2x

x^2 - 8 = 0

disc = b^2 - 4ac = 0 - 4 (1)(-8) = 32