Respuesta :
ANSWER
[tex]x=-2\pm 2\sqrt{2}[/tex]
EXPLANATION
Given:
[tex]x^2+4x-4=0[/tex]
We complete the square as follows;
Add 4 to both sides of the equation:
[tex]x^2+4x=4[/tex]
Add half the coefficient of [tex]x[/tex] to both sides of the equation.
[tex]x^2+4x+(2)^2=4+(2)^2[/tex]
The Left Hand Side of the equation is now a perfect square.
[tex](x+2)^2=4+4[/tex]
[tex](x+2)^2=8[/tex]
Take the square root of both sides
[tex]x+2=\pm \sqrt{8}[/tex]
[tex]x=-2\pm 2\sqrt{2}[/tex]
Hence the correct answer is option C
Given equation is [tex]x^2+4x-4=0[/tex]
compare with [tex]ax^2+bx+c=0[/tex]
we get: a=1, b=4, c=-4
plug these values into quadratic formula:
[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-4 \pm \sqrt{4^2-4(1)(-4)}}{2(1)}[/tex]
[tex]x=\frac{-4 \pm \sqrt{16-4(-4)}}{2}[/tex]
[tex]x=\frac{-4 \pm \sqrt{16+16}}{2}[/tex]
[tex]x=\frac{-4 \pm \sqrt{32}}{2}[/tex]
[tex]x=\frac{-4 \pm 4\sqrt{2}}{2}[/tex]
[tex]x=-2 \pm 2\sqrt{2}[/tex]
Hence finall answer is:
[tex]x=-2+2\sqrt{2}[/tex] and [tex]x=-2-2\sqrt{2}[/tex]