# Thread: Intersections of 2 lines

1. ## Intersections of 2 lines

Ok in this question, I can't draw graphs or use a calculator (I'm really wishing I could) but I need to find the intersection(s) of $y=\sqrt{x}$ and y=-x+2

I keep getting (1,4) or (1,2) but the answer is actually (1,1) so I'm not sure how to get to it?

2. In the problem, the equations are
y = sqrt(x) Which can be written as
y^2 = x.
Substitute this value in the second equation. You get
y = -y^2 + 2
Or
y^2 + y - 2 = 0
If you factorize yoy get
( y - 1 )( y + 2 ) = 0
Or y = -2 or y = 1.
Since y = sqrt(x), y cannot be negative.
So y = 1 and x = 1.

3. How did you get "(1, 4) or (1, 2)"? If x= 1 then y= -x+ 2= -1+ 2= 1, not either 4 or 2. Similarly, if x= 1, $y= \sqrt{x}= \sqrt{1}= 1$.

I would have done that a slightly different way. Since $y= \sqrt{x}$ and y= -x+ 2, $\sqrt{x}= -x+ 2$ (because they are both equal to y). Now square both sides. $x= (-x+ 2)^2= x^2- 4x+ 4$ so that $x^2- 5x+ 4= 0$. Notice that that is an equation for x where sa-ri-ga-ma got an equation for y.

$x^2- 5x+ 4= (x- 4)(x- 1)= 0$ gives x= 4 and x= 1. If x= 4, y= -x+ 2= -4+ 2= -2. But $y= \sqrt{x}$ cannot be negative so that is NOT a solution. If x= 1, y= -1+ 2= 1. The graphs intersect at (1, 1).

4. ## :D

Note the the line of function y=sqrt(x) OR $y= \sqrt{x}$ is "cut-off" at the y=0, it's because y can't be less than zero due to $y=\sqrt{x}$.