How would i find the discriminant and what type of solution would it be?

#a^2 + 4a + 4 = 0#

1 Answer
May 17, 2018

The standard form for a quadratic equation is:

#ax^2+ bx+c = 0#

where #x# is the independent variable and #a, b, and c# are constants.

The discriminant is:

#d = sqrt(b^2-4(a)(c))#

If #d < 0# then the quadratic equation has two complex conjugate roots.

If #d = 0# then the quadratic equation has one real root (Actually, it indicates that the quadratic is a perfect square and there are two real roots but they are the same value).

If #d > 0# then the quadratic equation has to distinct real roots.

Given #a^2 + 4a + 4 = 0#

Because your equation uses #a# ask the independent variable, we shall use #k# for the leading coefficient of the square term:

#d = sqrt(b^2-4(k)(c))#

Substitute the coefficients of the given equation, #k = 1, b=4 and c = 4#:

#d = sqrt(4^2-4(1)(4)#

#d = 0#

This is the case where the equation is a perfect square, therefore, there is only one real root.