Question

What values of x make the equation x^2+4x−4=0 true? Simplify any radicals

Answer 1

we can't factor so use quadratic formula

for
`ax^2+bx+c=0`
`x=(-b \pm √(b^2-4ac))/(2a)`

we have
`x^2+4x-4=0`

a=1, b=4, c=-4

`x=(-4 \pm √(4^2-4(1)(-4)))/(2(1))`

`x=(-4 \pm √(16+16))/(2)`

`x=(-4 \pm √(32))/(2)`

`x=(-4 \pm 4√(2))/(2)`

`x=-2 \pm 2√(2)`

x=-2+2√2 and x=-2-2√2 are the values of x that satisfy the equation

Answer 2

x^2 + 4x - 4 = 0

(x + 2)^2 - 4 - 4 = 0

(x + 2)^2 = 8

x + 2 = +/- √8

x = -2 + √8 , - 2 - √8

x = -2 + 2√2 , -2 - 2√2 (answer)