Question

A.

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.
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Answer 1

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

Explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

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) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\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) = (-3 + √(13))/(2*1) = 0.3`

`x_(2) = (-3 - √(13))/(2*1) = -3.3`

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