What are the zeros of the function f(x)=x^2+8x+4, expressed in the simplest radical form ?

Answer:[tex]x=-4 \pm 2\sqrt{3}[/tex]Step-by-step explanation:The zeros of f is when f=0.So we need to solve:[tex]x^2+8x+4=0[/tex]-----------------------------------------------------------------------------------------------------I'm going to choose completing the square.Subtract 4 on both sides:[tex]x^2+8x=-4[/tex]Add (8/2)^2 on both sides:[tex]x^2+8x+(\frac{8}{2})^2=-4+(\frac{8}{2})^2[/tex]Write left hand side as a square:[tex](x+\frac{8}{2})^2=-4+4^2[/tex][tex](x+4)^2=-4+16[/tex][tex](x+4)^2=12[/tex]Take the square root of both sides:[tex](x+4)=\pm \sqrt{12}[/tex][tex]x+4=\pm \sqrt{12}[/tex]Simplify right hand side:[tex]x+4=\pm \sqrt{4}\sqrt{3}[/tex][tex]x+4=\pm 2\sqrt{3}[/tex]Subtract 4 on both sides:[tex]x=-4 \pm 2\sqrt{3}[/tex]---------------------------------------------------------------------------------------------You could also go with quadratic formula:[tex]a=1[/tex][tex]b=8[/tex][tex]c=4[/tex]We need to use this formula:[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].I'm going to evaluate [tex]b^2-4ac[/tex] first.[tex]b^2-4ac=(8)^2-4(1)(4)=64-16=48[/tex]So now we have this so for with the formula:[tex]x=\frac{-8 \pm \sqrt{48}}{2(1)}[/tex]Simplifying the denominator gives:[tex]x=\frac{-8 \pm \sqrt{48}}{2}[/tex]Now is there a perfect square in 48?  Yes, 16 is a perfect square factor in 48.So we can write:[tex]x=\frac{-8 \pm \sqrt{16}\sqrt{3}}{2}[/tex][tex]x=\frac{-8 \pm 4\sqrt{3}}{2}[/tex][tex]x=\frac{-8}{2}\pm \frac{4\sqrt{3}}{2}[/tex]Reduce fractions:[tex]x=-4 \pm 2\sqrt{3}[/tex]