How do you find the discriminant of 7x^2+6x+2=07x2+6x+2=0 and use it to determine if the equation has one, two real or two imaginary roots?

2 Answers
Mar 18, 2017

7x^2+6x+27x2+6x+2 has complex number roots

Explanation:

Consider the standard format of y=ax^2+bx+cy=ax2+bx+c

Where x=(-b+-sqrt(b^2-4ac))/(2a)x=b±b24ac2a

The discriminant is the part of b^2-4acb24ac

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 <0b24ac=(6)24(7)(2)=3656<0

So 7x^2+6x+27x2+6x+2 has complex number roots

Mar 18, 2017

The solutions are S={-3/7+sqrt5/7i,-3/7-sqrt5/7i}S={37+57i,3757i}

Explanation:

The quadratic equation is

ax^2+bx+c=0ax2+bx+c=0

Here, we have

7x^2+6x+2=07x2+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