# Graph + Algebra Question

1. Sep 15, 2007

### tigerd12

1. The problem statement, all variables and given/known data

Hello, I am a 14 year old boy doing my GCSE's for Maths. And I was doing some past papers and I came across this question which I have never seen before. I looked in textbooks and found none of the sort, and I was wondering if any of you could please help :)

I have drawn the graph y=x+2/x

The solutions of the equation 2x+2/x=7 are the x-coordinates of the points of intersection of the graph of y=x+2/x and a straight line L.

Fine the Equation of L.

2. Relevant equations

None

3. The attempt at a solution

My guess at the answer would be the equation of the line is y=7 as 2x+2/x=7 means y=2x+2/x=7 as far as I remember. I drew this line on my graph, getting a solution of x=0.3, put this back into the equation and got an answer of 7.23(recurring). Which I thought was close enough allowing for errors, but I really have no idea if it's right or anything, it's just a guess.

Thanks in advance for any help :)

2. Sep 15, 2007

### chaoseverlasting

Actually, first you need to simplify the equation 2x+2/x=7 to $$2x^2-7x+2=0$$. This is a quadratic of the form $$ax^2+bx+c=0$$ where a=2, b=-7 and c=2.

The two points you're looking for can be found by $$x=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ and $$x=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

Next, you plug in the two values you get from here into the equation y=x+2/x to get two values of y.

Now, you have two points (x1,y1) and (x2,y2) (from your two values of x and y). Using these and the slope intercept form, you can find the equation of the line as such:

$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$

Last edited by a moderator: Sep 15, 2007
3. Sep 16, 2007

### tigerd12

Thanks!

Thanks alot! Just one question, the last formula you stated, I have never come across it before!! :P Mind explaining it?.. In the meantime, i'll try to figure it out myself, thanks alot though!

4. Sep 16, 2007

### HallsofIvy

What eqaution would you use for a straight line through two given points?

5. Sep 16, 2007

### tigerd12

I'm sorry, I don't know. As far as equations of straight lines goes, i've only ever had to use the basic primary school formula rise/run, which I know is also delta y/delta x. I think I am beginnning to get the solution though, I have plotted the two points on my graph and if I join them I get a straight line. Now when I do this, the gradient of the line is 2.3recurring, which I round off to 2 considering errors, and the Y intercept is about 7. So I think the solution may be y=2x+7.

6. Sep 16, 2007

### symbolipoint

Tigerd12, the straight line L would be some point (x, 7), which when you substitute y=7, would be the line which contains this point. You wrote:
You are then looking for the solution to a quadratic equation. The solutions will be the x-intercepts, the x coordinates for y=0 (relax; you already made the substitution for y=7 to obtain the equation. You just use algebra to put the equation into the form ax^2 + bx + c =0). Chaoseverlasting gave the right approach.

You will not really need the equation form for a line given in his message, but you likely need the knowledge as prerequisite development for the kind of question which you are working on; students learning about equations of lines before they learn about quadratic equations.