Using Bisection Method to Find Points of Intersection for y=x^3-2x+1 and y=x^2

gomes.
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y=x^3-2x+1

y=x^2


The question says to use the bisection method to find the points of intersection of the 2 curves. I know how to use the bisection method to find the root of an equation (like on this page http://kr.cs.ait.ac.th/~radok/math/mat7/step7.htm ), but how would I use it to find the point of intersection? Thanks.
 
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Well let (x_i,y_i) be a point of intersection. Can you use those curves to derive an equation for which either x_i or y_i is a root?
 
gomes. said:
question says to use the bisection method to find the points of intersection of the 2 curves. I know how to use the bisection method to find the root of an equation (like on this page http://kr.cs.ait.ac.th/~radok/math/mat7/step7.htm ), but how would I use it to find the point of intersection? Thanks.

What do you get if you subtract one equation from the other?
 
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thanks, I've solved it now :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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