Question

How do you find the discriminant of `7x^2+6x+2=0` and use it to determine if the equation has one, two real or two imaginary roots?

Answer 1

`7x^2+6x+2` has complex number roots

Explanation:

Consider the standard format of `y=ax^2+bx+c`

Where `x=(-b+-sqrt(b^2-4ac))/(2a)`

The discriminant is the part of `b^2-4ac`

If this is 0 then the vertex is actually on the on the axis so has 1 point

If this is negative then the roots are complex numbers

If this is positive and greater than 0 then it has two roots
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Added comment: I have had people argue that when the discriminant is 0 there is still 2 roots but they are the same value. This condition is called duplicity.
I do not like this!
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So for this equation:

Discriminant `-> b^2-4ac = (6)^2-4(7)(2) = 36 -56 <0`

So `7x^2+6x+2` has complex number roots

Answer 2

The solutions are `S={-3/7+sqrt5/7i,-3/7-sqrt5/7i}`

Explanation:

The quadratic equation is

`ax^2+bx+c=0`

Here, we have

`7x^2+6x+2=0`

The discriminant is

`Delta=b^2-4ac=36-4*7*2=36-56=-20`

As `Delta<0`, there are no real roots.

The roots are imaginary

`x=(-b+-sqrtDelta)/(2a)`

`x=(-6+-i2sqrt5)/(14)`

`x_1=-3/7+sqrt5/7i`

`x_2=-3/7-sqrt5/7i`