{3x}^(2) + 6x + 1 = 0 {3x}^(2) + 6x + 1 = 0 a, Are the roots equal or unequal? b. Are the roots rational or irrational? C. Are the roots real or non-real?​

Question

`{3x}^(2) + 6x + 1 = 0`


`{3x}^(2) + 6x + 1 = 0`

a, Are the roots equal or unequal? b. Are the roots rational or irrational? C. Are the roots real or non-real?​

Answer 1

• the roots are unequal (distinct)
• the roots are irrational
• the roots are real

Explanation:

A description of the roots of a quadratic can be had by looking at the value of the discriminant: b²-4ac, where a, b, c are the coefficients of the standard-form quadratic.

Here, we have a=3, b=6, c=1 so the discriminant is ...

d = 6² -4(3)(1) = 24

This is a positive integer that is not a perfect square. This value means ...

• the roots are unequal (distinct)
• the roots are irrational
• the roots are real

_____

The interpretation of the discriminant is ...

• d > 0, distinct real roots
• d = 0, one real root with multiplicity 2
• d < 0, complex roots

If d is an integer that is not a perfect square, the roots (or their imaginary part) will be irrational.