what are the zeros of the quadratic function $f(x)=6x^{2}+12x - 7$?\n$x=-1-sqrt{\frac{13}{6}}$ and…

what are the zeros of the quadratic function $f(x)=6x^{2}+12x - 7$?\n$x=-1-sqrt{\frac{13}{6}}$ and $x=-1+sqrt{\frac{13}{6}}$\n$x=-1-\frac{2}{sqrt{3}}$ and $x=-1+\frac{2}{sqrt{3}}$\n$x=-1-sqrt{\frac{7}{6}}$ and $x=-1+sqrt{\frac{7}{6}}$\n$x=-1-\frac{1}{sqrt{6}}$ and $x=-1+\frac{1}{sqrt{6}}$

what are the zeros of the quadratic function $f(x)=6x^{2}+12x - 7$?\n$x=-1-sqrt{\frac{13}{6}}$ and $x=-1+sqrt{\frac{13}{6}}$\n$x=-1-\frac{2}{sqrt{3}}$ and $x=-1+\frac{2}{sqrt{3}}$\n$x=-1-sqrt{\frac{7}{6}}$ and $x=-1+sqrt{\frac{7}{6}}$\n$x=-1-\frac{1}{sqrt{6}}$ and $x=-1+\frac{1}{sqrt{6}}$

Answer

Explanation:

Step1: Recall quadratic - formula

For a quadratic function $f(x)=ax^{2}+bx + c$, the zeros are given by $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 6$, $b = 12$, and $c=-7$.

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

Substitute the values of $a$, $b$, and $c$ into the discriminant formula: $\Delta=(12)^{2}-4\times6\times(-7)=144 + 168=312$.

Step3: Apply the quadratic - formula

$x=\frac{-12\pm\sqrt{312}}{2\times6}=\frac{-12\pm2\sqrt{78}}{12}=\frac{-6\pm\sqrt{78}}{6}=\frac{-6\pm\sqrt{6\times13}}{6}=\frac{-6\pm\sqrt{6}\times\sqrt{13}}{6}=-1\pm\frac{\sqrt{13}}{\sqrt{6}}=-1\pm\sqrt{\frac{13}{6}}$.

Answer:

A. $x=-1-\sqrt{\frac{13}{6}}$ and $x=-1 + \sqrt{\frac{13}{6}}$