Question

X^2+x-4

`{x}^(2) + 4x - 4 = 0`

Answer 1

`x^2+4x-4=0\qquad\text{add 8 to both sides}\\\\x^2+4x+4=8\\\\x^2+2(x)(2)+2^2=8\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+2)^2=8\iff x+2=\pm\sqrt8\\\\x+2=-√(4\cdot2)\ \vee\ x+2=√(4\cdot2)\\\\x+2=-\sqrt4\cdot\sqrt2\ \vee\ x+2=\sqrt4\cdot\sqrt2\\\\x+2=-2\sqrt2\ \vee\ x+2=2\sqrt2\qquad\text{subtract 2 frim both sides}\\\\\boxed{x=-2-2\sqrt2\ \vee\ x=-2+2\sqrt2}`