what are the zeros of the function $f(x)=x^{2}+8x + 4$, expressed in simplest radical form?\n$x=-4pm2sqrt{3}$…

what are the zeros of the function $f(x)=x^{2}+8x + 4$, expressed in simplest radical form?\n$x=-4pm2sqrt{3}$\n$x=-4pmsqrt{48}$\n$x=\frac{-8pmsqrt{3}}{2}$\n$x=\frac{-4pm4sqrt{3}}{2}$

what are the zeros of the function $f(x)=x^{2}+8x + 4$, expressed in simplest radical form?\n$x=-4pm2sqrt{3}$\n$x=-4pmsqrt{48}$\n$x=\frac{-8pmsqrt{3}}{2}$\n$x=\frac{-4pm4sqrt{3}}{2}$

Answer

Explanation:

Step1: Recall quadratic - formula

For a quadratic function (y = ax^{2}+bx + c), the zeros are given by (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}). In the function (f(x)=x^{2}+8x + 4), (a = 1), (b = 8), and (c = 4).

Step2: Calculate the discriminant (\Delta=b^{2}-4ac)

Substitute (a = 1), (b = 8), (c = 4) into (\Delta). (\Delta=8^{2}-4\times1\times4=64 - 16=48).

Step3: Find the zeros using the quadratic - formula

(x=\frac{-8\pm\sqrt{48}}{2\times1}=\frac{-8\pm4\sqrt{3}}{2}=-4\pm2\sqrt{3}).

Answer:

(x=-4\pm2\sqrt{3}) (corresponds to the first option (x=-4\pm2\sqrt{3}))