determine the zeros of the function $f(x)=3x^{2}-7x + 1$.\n$x=\frac{7pmsqrt{37}}{6}$\n$x=\frac{-7pmsqrt{37}}{…

determine the zeros of the function $f(x)=3x^{2}-7x + 1$.\n$x=\frac{7pmsqrt{37}}{6}$\n$x=\frac{-7pmsqrt{37}}{6}$\n$x=\frac{7pmsqrt{61}}{6}$\n$x=\frac{-7pmsqrt{61}}{6}$

determine the zeros of the function $f(x)=3x^{2}-7x + 1$.\n$x=\frac{7pmsqrt{37}}{6}$\n$x=\frac{-7pmsqrt{37}}{6}$\n$x=\frac{7pmsqrt{61}}{6}$\n$x=\frac{-7pmsqrt{61}}{6}$

Answer

Explanation:

Step1: Recall quadratic - formula

For a quadratic function $ax^{2}+bx + c = 0$, the solutions are given by $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In the function $f(x)=3x^{2}-7x + 1$, we have $a = 3$, $b=-7$, and $c = 1$.

Step2: Calculate the discriminant

The discriminant $\Delta=b^{2}-4ac$. Substitute $a = 3$, $b=-7$, and $c = 1$ into the formula: $\Delta=(-7)^{2}-4\times3\times1=49 - 12=37$.

Step3: Find the zeros

Using the quadratic - formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, substitute $a = 3$, $b=-7$, and $\Delta = 37$: $x=\frac{-(-7)\pm\sqrt{37}}{2\times3}=\frac{7\pm\sqrt{37}}{6}$.

Answer:

$x=\frac{7\pm\sqrt{37}}{6}$