Reversing?? an equation

[Jessica]

y = x*(x-1)*5;

 

Given $y, calculate $x.

 

I can get this far, it's been so long since I took a math class that I'm not even sure this makes sense.

y/5 = x*(x+1)

(y/5)/x = x+1

((y/5)/x)-1 = x

 

My problem is: Given $y = 60, how do I get $x = 4 (using php)

 

60 = x*(x-1)*5;

12 = x*(x-1);

12/x = x-1

(12/x)+1 = x

 

And I'm stuck. (I know $x = 4 because I made a spreadsheet plugging in 1-100 as x and then getting y, so where x = 4 y = 60)

[xyph]

I'd tell you to expand the brackets, then solve, but we're on the internet here.

 

http://www.wolframalpha.com/input/?i=y+%3D+x*%28x-1%29*5

 

-5x²+5x+y = 0

 

[Jessica]

I don't get how that gives me x?

 

I see on wolfram alpha the two alternate forms but neither gives you x when you know y

 

I want to write a calculation in PHP that if I say I have 60 for y, it can tell me x = 4. Or y=100, and it tells me x = 5. And so on. If I give it 80 it should calculate for x and then round down to 4 but I can handle that part

[xyph]

x will have two values for any given y, generally.

[Jessica]

Ah, I get it. hence the parabola. Thanks

Is there any mathematical way to determine the positive value? Probably not...

[kicken]

http://www.wolframalpha.com/input/?i=y+%3D+x*%28x-1%29*5+solve+for+x

 

Neat tool that wolfram alpha is.  I should use that more.

[Jessica]

http://www.wolframalpha.com/input/?i=y+%3D+x*%28x-1%29*5+solve+for+x

 

Neat tool that wolfram alpha is.  I should use that more.

 

Seriously. I am bookmarking that site, I'd heard of it but never used it before. You guys rock.

 

So let's see in PHP this would be...

 

<?php
$y = 60;
$x = .1*((sqrt(5)*sqrt((4*$y)+5))+5);
echo $x;
?>

 

And it echoed 4. Perfect!!

 

 

[phpdevelopers]

It was really a helpful content. Keep sharing…

[ManiacDan]

If you don't want to be big fat cheaters, the way to do this on paper is to use the quadratic formula For this equation, it's as easy as:

 

 

y = x*(x-1)*5;

 

y = 5x^2 - 5x; //jesi, you messed up here by moving X back to both sides of the equation. Don't do that

 

0 = 5x^2 - 5x - y; //everything has to equal zero

 

//here, we substitute Jesi's value for Y and use the Quadratic Formula

 

x = (5 +/- sqrt(25 - 4 * 5 * -60)) / 10;

 

x = (5 +/- 35) / 10;

 

x = 4,3;

 

The parabola formed by the equation cuts the Y axis at 4 and 3. Both are valid answers, but only the larger one satisfies the original equation.

 

Also, I just now noticed this thread is 6 months old and newbie here bumped it, but I did all the math so you're getting the post anyway.