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

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

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 = (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 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) = (-(-2) + √(64))/(2*3) = 1.67`

`x_(2) = (-(-2) - √(64))/(2*3) = -1`

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