Expert: Steve Holleran - 4/2/2008

Question
I am attempting to finish homework problems but I have difficulty with set up can you assist me?
1.2x^2+5x+5=0
2.p^2-4p+4=0
3.m^2+m+1=0
43z^2+2-1=0
5. What type of solution do you get for quadratic equations where D<0. I have to give reason for my answer, and also give an example with the solution.

Answer
Hi Audrey,

It looks like, in #1 - #4, you have a mixture of factoring and using the quadratic formula.

Let's take the ones that factor first: 2 and 4:

#2) p^2 - 4p + 4 = 0---->  (p - 2)(p-2) = 0 so p = 2 is a double root

#4)  3z^2 + 2z - 1 = 0 --->  (3z - 1)(z + 1) = 0

                             so z = 1/3 or z = -1.

#1 and #3 have to be solved by the quadratic formula:

      x = [-b +/- sqrt(b^2 - 4ac)]/2a

1)  a = 2, b = 5 and c = 5,  so x = [-5 +/- sqrt(-15)]/4

   or,  x = -5 +/- i * sqrt(15)      where i is the imaginary unit

3)  a = 1, b = 1, c = 1, so  x = [-1 +/1 sqrt(-3)]/2

   or, x = [-1 +/- i * sqrt(3)]/2 or x = -1/2 +/- i * sqrt(3)/2

For #5, The D referred to is the quantity under the square root sign, called the Discriminant.  If D < 0, you can see from the solutions to #1 and #3 that you get imaginary (complex) solutions, because you are taking square roots of negative numbers.

#3 is a very good example of this.  Note also that this means that the graphs of equations like #1 and #3 will not cross the x-axis.  If they had real solutions, like #2 and #4, they would.

Steve