SOLUTION: How many real number solutions does the equation y=2x^2−8x+8 have?

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Question 1181964: How many real number solutions does the equation y=2x^2−8x+8 have?
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
How many real number solutions does the equation y=2x%5E2-8x%2B8 have?
use discriminant b%5E2-4ac
if b%5E2-4ac%3E0, the equation y=2x%5E2-8x%2B8 will have two distinct real number solutions
in your case a=2, b=-8, and c=8
substitute in b%5E2-4ac%3E0
%28-8%29%5E2-4%2A2%2A8%3E0
64-64%3E0
0%3E0 -> false => the equation y=2x%5E2%E2%88%928x%2B8 have no two distinct real number solutions
if b%5E2-4ac=0, the equation y=2x%5E2-8x%2B8 will have a repeated real number solution
%28-8%29%5E2-4%2A2%2A8=0
64-64=0
0=0 ->true=> the equation y=2x%5E2-8x%2B8 will have onea repeated real number solution

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The question you ask is undoubtedly NOT the question you intended to ask.

The equation y=2x^2-8x+8 has an infinite number of real number solutions. Pick any number you want for x and plug into the equation to find y.

(x,y) = (0,8) is a solution
(x,y) = (1,2) is a solution
(x,y) = (sqrt(3),14-8*sqrt(3)) is a solution

and so on, ad infinitum....

The question you INTENDED to ask was either "how many distinct roots does the equation y=2x^2-8x+8 have?" or "how many solutions does the equation 2x^2-8x+8=0 have?"

The answer to that question is "one".

2x^2-8x+8=0
x^2-4x+4=0
(x-2)(x-2)=0

The equation has one real number solution: x=2.

The equation is quadratic, so it has two roots; but the two roots are the same, so it has one distinct root.

In summary, using correct mathematical language....

The equation y=2x^2-8x+8 has one distinct real root

The equation 2x^2-8x+8=0 has one real solution

The equation y=2x^2-8x+8 has an infinite number of real solutions