Intersection of a line and a curve

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Just wondering if you could check my working for me below. I *think* it's right but when I put it in on an equation calculator to check and it seems to think that x is just 0 but that might just be it only giving one value (origin). Regardless, I'm mainly interested in whether I re-arranged it for x correctly and factorised it correctly, not just the answer.

Question:

Find the point of intersection of the curve y = sqr.x and y = 1/2x other than the origin.

Working:

Equate both formulas:

1/2x = sqr.x

Square both sides:

1/4x^2 = x

Move all x to one side:

0 = x - 1/4x^2

Factorise x out:

0 = x(1-1/4x)

----

x = 0, but other than the origin, the othe point of intersection is:

(1-1/4x) = 0
1 = 1/4x

Multiply by 4:

4 = x
x = 4

----

When x = 4:

y = sqr.x
y = sqr.4
y = 2

Therefore point = (4,2)

This was just the start of the question for finding the area bound by the curve and the line which I have done, but was unsure here (on the easier bit hehe :D) for some odd reason.

Anyway, thanks for any help. I know it can also factorise to x(x-4) if you do the steps a bit differently at the start by doubling to get 1 whole x, squaring both sides and then re-arranging. That also gives 4 but I just wanted to check.

Yeah it's a bit basic, sorry, but thanks for checking. Oh and by sqr. I mean square root if that's not clear.

Not sure if this was the right board on here to post it on but cheers. :)
 
Hello, Monkeyseat!

Your work and your reasoning is correct . . . Nice going!

And thank you for showing your work . . . a rare treat.

 
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