2. Re: equation

Originally Posted by Mhmh96

According to wolframalpha your equation does not have real solutions, so you have to find such a way to prove that.

3. Re: equation

This is tricky equation and simply making algebraic manipulations will require a lot of work.

In fact this equation doesn't have real roots.
Let's see why.

First of all domain of this equation is x>0.
Next. It is known that for a>0 and b>0 $a+b \geq 2\sqrt{ab}$
Proof is straightforward. $(\sqrt{a}-\sqrt{b})^2\geq 0$ -> $a+b- 2\sqrt{ab}\geq 0$ -> $a+b \geq 2\sqrt{ab}$.

Now,
$\frac {3x+3}{\sqrt{x}}=3\left(\sqrt{x}+\frac{1}{\sqrt{x} }\right) \geq 3 \left(2\sqrt{\sqrt{x}*\frac{1}{\sqrt{x}}}\right)=6$
Therefore, left side is greater or equal than 6 for all x>0.

Now, let's see how to handle right part.

Take expression $5x^2-9x+9$ and complete the square: $5x^2-9x+9=\left(\sqrt{5}x-\frac{9}{2\sqrt{5}}\right)^2+9-\frac{81}{20}>0$
From this we have that $5x^2-9x+9>0$ for all x.
Now, divide this expression by 4: $1.25x^2-2.25x+2.25>0$ or $2.25x^2-2.25x+2.25>x^2$.
Finally we have that $2.25(x^2-x+1)>x^2$
Dividing by $x^2-x+1$ (it is positive for all values of x) gives $2.25>\frac{x^2}{x^2-x+1}$
Taking square roots and remembering that x>0 yields
$\sqrt{\frac{x^2}{x^2-x+1}}<1.5$ or $\frac{x}{\sqrt{x^2-x+1}}<1.5$

Thus, $4+\frac{x}{\sqrt{x^2-x+1}}<4+1.5=5.5$

So, what did we show?
We showed that left part is at least 6 and right part is less than 5.5. There are no intersections, so equation has no real roots.

4. Re: equation

According to WolframAlpha, the RHS attains a maximal value of $4 + \frac{2}{\sqrt{3}}$. Still less than 6...

Proving this using calculus is quite trivial...not a lot of easy algebraic solutions.

5. Re: equation

Actually ,i meant ,i dropped 1 so there is no real solution for the previous one .

6. Re: equation

Originally Posted by Mhmh96
Actually ,i meant ,i dropped 1 so there is no real solution for the previous one .
Here is WA's solution.

7. Re: equation

See, that's why you check what you post before you post it.

8. Re: equation

Thats right richard1234 ,i know that the answer is 1,but is there a particular way to solve it ?

9. Re: equation

Best bet would be to assume that $x = \tan^2 \theta$ (because if x is real, then $x \ge 0$).

Other than that, I don't see any other potentially easy solutions.

WolframAlpha says the only real solution is $x = 1$.