# A.Approximate the real zeros of f(x) = 3x + x2 - 1 to the nearest tenthA. -0.7,0.7B. -7,7b. -0.5, 0.5d.

###### Question:

Approximate the real zeros of f(x) = 3x + x2 - 1 to the nearest tenth

A. -0.7,0.7

B. -7,7

b. -0.5, 0.5

d. -0.6, 0.6

Please select the best answer from the choices provided.

## Answers

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.