Answer:
The real zeros of f(x) are x = 0.3 and x = -3.3.
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
In this problem, we have that:
[tex]f(x) = x^{2} + 3x - 1[/tex]
So
[tex]a = 1, b = 3, c = -1[/tex]
[tex]\bigtriangleup = 3^{2} - 4*1*(-1) = 13[/tex]
[tex]x_{1} = \frac{-3 + \sqrt{13}}{2*1} = 0.3[/tex]
[tex]x_{2} = \frac{-3 - \sqrt{13}}{2*1} = -3.3[/tex]
The real zeros of f(x) are x = 0.3 and x = -3.3.
