What Is the Equation of Line L Intersecting the Graph y=x+2/x?

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To find the equation of line L intersecting the graph y=x+2/x, the equation 2x+2/x=7 simplifies to the quadratic 2x^2-7x+2=0. Solving this quadratic gives the x-coordinates of the intersection points. These x-values are then substituted back into the original equation to find corresponding y-values. The slope of the line can be determined from these points, leading to the final equation of the line, which is approximated as y=2x+7. Understanding the relationship between the quadratic and linear equations is essential for solving this problem.
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Homework Statement



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.

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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 :)
 
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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)
 
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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 a lot though!
 
What eqaution would you use for a straight line through two given points?
 
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.
 
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:
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.
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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