Question

PLLLLZZZ HELP ME. Given the function f(x)=3x^2-2x-5 : What are the zeros for this function? 5 extra credit points for an exact answer. 30 points if you get the exact answer and its 100% Correct.

Answer 1

Answer:

The zeros for this function are x = -1 and x = 1.67

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 $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, 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 question:

$f(x) = 3x^{2} - 2x - 5$

The zeros of the function are the values of x for which

$f(x) = 0$

Then

$3x^{2} - 2x - 5 = 0$

This means that $a = 3, b = -2, c = -5$

Then

$\bigtriangleup = (-2)^{2} - 4*3*(-5) = 64$

$x_{1} = \frac{-(-2) + \sqrt{64}}{2*3} = 1.67$

$x_{2} = \frac{-(-2) - \sqrt{64}}{2*3} = -1$

The zeros for this function are x = -1 and x = 1.67