# Silvanus P. Thompson Text problem

• May 31st 2008, 09:33 PM
Haakon
Silvanus P. Thompson Text problem
Hi there,

I am a Physics teacher and for a bit of fun I am working through Calculus made easy by Thompson. I have got stuck on some algebra on q.7 exercise X. The question asks " Find the maxima and minima of y = 3x /(x^2 - 3) + 0.5x + 5.
So I have differentiated this and got (-3x^2 - 9)/(x^2-3)^2 + 5

I then equated the differential to zero and tried to solve for x. I got the equation down to 12x^2 + 9 = x^4 but now I am stuck.. How do I simplify this to get x? http://www.mathhelpforum.com/math-he...cons/icon5.gif
• May 31st 2008, 09:48 PM
Chris L T521
Quote:

Originally Posted by Haakon
Hi there,

I am a Physics teacher and for a bit of fun I am working through Calculus made easy by Thompson. I have got stuck on some algebra on q.7 exercise X. The question asks " Find the maxima and minima of y = 3x /(x^2 - 3) + 0.5x + 5.
So I have differentiated this and got (-3x^2 - 9)/(x^2-3)^2 + 5

I then equated the differential to zero and tried to solve for x. I got the equation down to 12x^2 + 9 = x^4 but now I am stuck.. How do I simplify this to get x? http://www.mathhelpforum.com/math-he...cons/icon5.gif

Hello Haakon!

I would recommend writing it as a quartic equation: $x^4-12x^2-9=0$. Now make the substitution $z=x^2$. This substitution lets us solve the quadratic equation $z^2-12z-9=0$. Solving for z, we get:

$z=\frac{12\pm\sqrt{144+36}}{2}\implies z=6\pm 3\sqrt{5}$.

But $z=x^2$, thus $x^2=6\pm 3\sqrt{5}\implies x=\pm\sqrt{6\pm 3\sqrt{5}}$.

Hope this made sense! :D