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

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) (Show Source): You can put this solution on YOUR website!
How many real number solutions does the equation have?
use discriminant
if , the equation will have two distinct real number solutions
in your case , , and
substitute in

-> false => the equation have no two distinct real number solutions
if , the equation will have a repeated real number solution

->true=> the equation will have a repeated real number solution

Answer by greenestamps(13198) (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

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