SOLUTION: x^2-8x+16=0. Determine the type (real or complex) and number of the solutions of each of the following equations. Justify your answer. Thank You

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: x^2-8x+16=0. Determine the type (real or complex) and number of the solutions of each of the following equations. Justify your answer. Thank You      Log On


   

Question 194140: x^2-8x+16=0.
Determine the type (real or complex) and number of the solutions of each of the following equations. Justify your answer.
Thank You
Found 2 solutions by josmiceli, jim_thompson5910:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square:
Take 1/2 of the coefficient of x
square it, and add to both sides
x%5E2+-+8x+%2B+16+=+0
First, subtract 16 from both sides
x%5E2+-+8x+=+-16
x%5E2+-+8x+%28-8%2F2%29%5E2+=+-16+%2B+%28-8%2F2%29%5E2
x%5E2+-+8x+%2B+16+=+-16+%2B+16
The left side is now a perfect square
%28x+-+4%29%5E2+=+0
%28x+-+4%29%28x+-+4%29+=+0
This has double roots, and for each one,
x+=+4 which is real, so 2 real roots

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
From x%5E2-8x%2B16 we can see that a=1, b=-8, and c=16


D=b%5E2-4ac Start with the discriminant formula.


D=%28-8%29%5E2-4%281%29%2816%29 Plug in a=1, b=-8, and c=16


D=64-4%281%29%2816%29 Square -8 to get 64


D=64-64 Multiply 4%281%29%2816%29 to get %284%29%2816%29=64


D=0 Subtract 64 from 64 to get 0


Since the discriminant is equal to zero, this means that there is one real solution.