SOLUTION: Determine how many real roots the equation has (x-4)^2=0 I know there are 1, which means when i put it in b^2-4ac it equals 0, but i dont know how to get it into that form.

Algebra ->  Linear-equations -> SOLUTION: Determine how many real roots the equation has (x-4)^2=0 I know there are 1, which means when i put it in b^2-4ac it equals 0, but i dont know how to get it into that form.      Log On


   

Question 111499This question is from textbook
: Determine how many real roots the equation has (x-4)^2=0
I know there are 1, which means when i put it in b^2-4ac it equals 0, but i dont know how to get it into that form. This question is from textbook

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-4%29%5E2=0
x%5E2+-+8x+%2B+16+=+0
this is in the form ax%5E2+%2B+bx+%2B+c+=+0
a = 1
b = -8
c = 16
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
As long as b%5E2+-+4ac is a positive number, the roots are real
If b%5E2+=+4ac, there is a single real root
b%5E2+-+4ac+=+%28-8%29%5E2+-+4%2A1%2A16
b%5E2+-+4ac+=+64+-+64
b%5E2+-+4ac+=+0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-8%29+%2B-+sqrt%280+%29%29%2F%282%2A1%29+
x+=+8%2F2
x+=+4 answer