Solve (ab)^2 + ab - a^2 - 4b^2 = 0

  • Thread starter Thread starter soopo
  • Start date Start date
soopo
Messages
222
Reaction score
0

Homework Statement



Solve the following equation for `a`.
(ab)^2 + ab - a^2 - 4b^2 = 0


The Attempt at a Solution



ab( ab + 1 ) - a^2 -4b^2 =

I perhaps should now use some sort of Diophantine equation.
However, I am not sure.
 
Physics news on Phys.org
Just treat it as a quadratic equation in 'a', regarding 'b' as a constant.
 
hiya,

treat b like a constant.
maybe the best way to solve this for a is to put the equation in the form of a quadratic, like:
ax^2+bx+c=0
(in this case you will have a instead of x) then use the quadratic formula to find the value(s) for a.
 
Dick said:
Just treat it as a quadratic equation in 'a', regarding 'b' as a constant.

sara_87 said:
hiya,

treat b like a constant.
maybe the best way to solve this for a is to put the equation in the form of a quadratic, like:
ax^2+bx+c=0
(in this case you will have a instead of x) then use the quadratic formula to find the value(s) for a.
Two great minds with a single thought!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top