Question

Solve 3x^2 + x + 10 = 0 round solutions to the nearest hundredth

A. X= -2.83 and x=0.83

B. No real solutions

C. X= -2.01 and x= 1.67

D. X= -1.67 and x=2.01

Answer 1

C. X= -2.01 and x= 1.67

Explanation:

`3x {}^(2) + x + 10 = 0 \\ 3x {}^(2) + 6x - 5x + 10 = 0 \\ 3x(x + 2) - 5(x + 2) \\ (x + 2)(3x - 5 )\\ x + 2 = 0 \: \: or \: \: 3x - 5 = 0 \\ x = - 2 \: \: or \: \: x = (5)/(3)`

Answer 2

B. No real solutions

EXPLANATION

The given equation is

`3 {x}^(2) + x + 10 = 0`

By comparing to

`a {x}^(2) + bx + c= 0`

We have a=3,b=1 and c=10.

We substitute these values into the formula

`D = {b}^(2) - 4ac`

to determine the nature of the roots.

`D = {1}^(2) - 4(3)(10)`

`D = 1 - 120`

`D = - 119`

The discriminant is negative.

This means that the given quadratic equation has no real roots.