what are the zeros of the quadratic function $f(x)=2x^{2}-10x - 3$?\n$x =-\frac{5}{2}-\frac{sqrt{31}}{2}$…

what are the zeros of the quadratic function $f(x)=2x^{2}-10x - 3$?\n$x =-\frac{5}{2}-\frac{sqrt{31}}{2}$ and $x =-\frac{5}{2}+\frac{sqrt{31}}{2}$\n$x =-\frac{5}{2}-sqrt{\frac{37}{8}}$ and $x =-\frac{5}{2}+sqrt{\frac{37}{8}}$\n$x =\frac{5}{2}-\frac{sqrt{31}}{2}$ and $x =\frac{5}{2}+\frac{sqrt{31}}{2}$\n$x =\frac{5}{2}-sqrt{\frac{37}{8}}$ and $x =\frac{5}{2}+sqrt{\frac{37}{8}}$

what are the zeros of the quadratic function $f(x)=2x^{2}-10x - 3$?\n$x =-\frac{5}{2}-\frac{sqrt{31}}{2}$ and $x =-\frac{5}{2}+\frac{sqrt{31}}{2}$\n$x =-\frac{5}{2}-sqrt{\frac{37}{8}}$ and $x =-\frac{5}{2}+sqrt{\frac{37}{8}}$\n$x =\frac{5}{2}-\frac{sqrt{31}}{2}$ and $x =\frac{5}{2}+\frac{sqrt{31}}{2}$\n$x =\frac{5}{2}-sqrt{\frac{37}{8}}$ and $x =\frac{5}{2}+sqrt{\frac{37}{8}}$

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)=2x^{2}-10x - 3), we have (a = 2), (b=-10), and (c=-3).

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

Substitute (a = 2), (b=-10), and (c=-3) into the discriminant formula. (\Delta=(-10)^{2}-4\times2\times(-3)=100 + 24=124).

Step3: Find the zeros using the quadratic - formula

[ \begin{align*} x&=\frac{-(-10)\pm\sqrt{124}}{2\times2}\ &=\frac{10\pm\sqrt{4\times31}}{4}\ &=\frac{10\pm2\sqrt{31}}{4}\ &=\frac{5\pm\sqrt{31}}{2} \end{align*} ] So the zeros are (x=\frac{5}{2}-\frac{\sqrt{31}}{2}) and (x=\frac{5}{2}+\frac{\sqrt{31}}{2}).

Answer:

(x=\frac{5}{2}-\frac{\sqrt{31}}{2}) and (x=\frac{5}{2}+\frac{\sqrt{31}}{2})