Problem
Find the zeros by using the quadratic formula and tell whether the solutions are real or imaginary. F(x)=x^2–2x–14.
Solution
For this case we can do the following:
x^2–2x–14.
We can use the quadratic formula given by:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where a =1, b= -2, c= -14
And replacing we got:
[tex]x=\frac{2\pm\sqrt[]{4-(4)(1)(-14)}}{2\cdot1}[/tex]And solving we got:
[tex]x=\frac{2\pm\sqrt[]{60}}{2}=1\pm\sqrt[]{15}[/tex]And the two solutions are:
x1= 1 +sqrt(15)
x2= 1- sqrt(15)
