Asked by Lindsay
I'm trying to find the intersection point of y=x-2 and y=sqrt(x). I tried letting both equations equal each other but kept coming up with the points (4,1) and (1,4) however these aren't right as the answer should be (4,2) but I am unable to get this answer. Help please?
Answers
Answered by
Reiny
so x-2 = √x
square both sides ...
x^2 - 4x + 4 = x
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x = 1 or x = 4
since we squared, both answers have to be verified in the original equation
if x = 1
LS = 1-2 = -1
RS = √1 = 1 ≠ LS
if x=4
LS = 4-2 = 2
RS = √4 = 2
so x = 4 is the only solution
if x = 4,
y = 4-2 = 2
They intersect at (4,2)
look at how Wolfram shows it
http://www.wolframalpha.com/input/?i=x-2+%3D+√x
square both sides ...
x^2 - 4x + 4 = x
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x = 1 or x = 4
since we squared, both answers have to be verified in the original equation
if x = 1
LS = 1-2 = -1
RS = √1 = 1 ≠ LS
if x=4
LS = 4-2 = 2
RS = √4 = 2
so x = 4 is the only solution
if x = 4,
y = 4-2 = 2
They intersect at (4,2)
look at how Wolfram shows it
http://www.wolframalpha.com/input/?i=x-2+%3D+√x