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?

2 Answers
Mar 18, 2017

Answer:

#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

Mar 18, 2017

Answer:

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#