SOLUTION: Which describes the number and type of roots of the equation x^2 + 121x = 0 a) 1 real root, two imaginary roots b) 2 real roots. 1 imaginary roots c) 3 real roots d) 3 imagin

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Which describes the number and type of roots of the equation x^2 + 121x = 0 a) 1 real root, two imaginary roots b) 2 real roots. 1 imaginary roots c) 3 real roots d) 3 imagin      Log On


   

Question 296641: Which describes the number and type of roots of the equation x^2 + 121x = 0
a) 1 real root, two imaginary roots
b) 2 real roots. 1 imaginary roots
c) 3 real roots
d) 3 imaginary roots
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
This quadratic equation has 2 real roots 0 and -121
Imaginary roots always come in pairs.
So b and d are out
I have already shown two real roots so a is out
The only possible answer among the choices is c
but a quadratic (x^2) can have a maximum of 2 roots.
So none of the choices are correct. There must be a misprint in the choices or the problem.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B121x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28121%29%5E2-4%2A1%2A0=14641.

Discriminant d=14641 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-121%2B-sqrt%28+14641+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28121%29%2Bsqrt%28+14641+%29%29%2F2%5C1+=+0
x%5B2%5D+=+%28-%28121%29-sqrt%28+14641+%29%29%2F2%5C1+=+-121

Quadratic expression 1x%5E2%2B121x%2B0 can be factored:
1x%5E2%2B121x%2B0+=+1%28x-0%29%2A%28x--121%29
Again, the answer is: 0, -121. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B121%2Ax%2B0+%29