Question

A.

Answer 1

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:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem, we have that:

$f(x) = x^{2} + 3x - 1$

So

$a = 1, b = 3, c = -1$

$\bigtriangleup = 3^{2} - 4*1*(-1) = 13$

$x_{1} = \frac{-3 + \sqrt{13}}{2*1} = 0.3$

$x_{2} = \frac{-3 - \sqrt{13}}{2*1} = -3.3$

The real zeros of f(x) are x = 0.3 and x = -3.3.