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# Algebra

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What are the coordinates of the points where the graphs  of f(x)=x^3 + x^2 - 3x + 5 and g(x) = x^3 + 2x^2 intersect?

Give your answer as a list of points separated by commas, with the points ordered such that their -coordinates are in increasing order. (So "(1,-3), (2,3), (5,-7)" - without the quotes - is a valid answer format.)

May 3, 2024

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Alright! I geuss I can try to solve this question!

First, let's note that the only points where the lines intersect is when we have $$f(x) = g(x)$$

This means that we only have to solve the equation $$x^3 + x^2 -3x + 5 = x^3 + 2x^2$$ to find the points where the lines intersect.

Combining like terms and moving all the terms to one side, we get the equation $$x^2 + 3x - 5 = 0$$.

We can't factor this polynomial with integers, so the next best thing to do is to complete the square and find x that way.

We set up the equation by moving 5 to the other side, getting us $$x^2 + 3x = 5$$. Now we complete the square by adding 9/4 to both sides.

$$x^2 + 3x + 9/4 = 5 + 9/4$$

$$(x + 3/2)^2 = 29/4$$

Now we just have to square root both sides to get us $$x+3/2= \pm \sqrt{29/4}$$

This means that $$x = \frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}$$.

Subsituting these x values back into the g(x) and f(x) functions, we get $$y = \frac{-\sqrt{29}-3} 2, \frac{\sqrt{29}-3}{2}$$

So for the final answer, we have $$(\frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}), (-\frac{\sqrt{29}-3}{2}, \frac{\sqrt{29}-3}{2})$$

Thanks! :) :) :)

May 3, 2024