Algebraic Methods for Solving Complicated Equations with Variable x

  • Thread starter Thread starter larry91
  • Start date Start date
larry91
Messages
12
Reaction score
0
Hi everybody!

I have to solve the following equation using algebraic methods: a=? such as 2x + ax > 3x + 4x for any x from ℝ.

I solved it using analytical methods. Using The Rolle Theorem, but I don't want to solve it in that way! The solution for a is 6
 
Physics news on Phys.org
larry91 said:
Hi everybody!

I have to solve the following equation using algebraic methods: a=? such as 2x + ax > 3x + 4x for any x from ℝ.

I solved it using analytical methods. Using The Rolle Theorem, but I don't want to solve it in that way! The solution for a is 6

Be careful: as written, your inequality fails for the real value x = 0. Perhaps you meant "≥". (In Mathematics, we need to be precise!) Also, of course, a solution is _any_ a ≥ 6. Perhaps the question asked for the smallest such a?

Finally, why do you not want to use Rolle's theorem?

RGV
 
Ray Vickson said:
Be careful: as written, your inequality fails for the real value x = 0. Perhaps you meant "≥". (In Mathematics, we need to be precise!) Also, of course, a solution is _any_ a ≥ 6. Perhaps the question asked for the smallest such a?

Finally, why do you not want to use Rolle's theorem?

RGV

Yes! The smallest value for a! and there is ≥. Sorry!
Using Rolle's theorem is quite simple to solve it... I don't want to solve it 'cause I need a solution for it... I just want to find an approach through we can solve it using algebraic methods!
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top