1)Solve this equation: x+√(x^2+4/x^2 +4)=4
2)a, b and c are the sides of a triangle and p , q are positive numbers and p+q = 1 . Prove that pa^2+qb^2≥pqc^2.
3)a and b are real positive numbers. a^n=a+1 and b^2n=b+3
n ∈N and n >1. Compare a with b.

2. Originally Posted by blertta
1)Solve this equation: x+√(x^2+4/x^2 +4)=4
2)a, b and c are the sides of a triangle and p , q are positive numbers and p+q = 1 . Prove that pa^2+qb^2≥pqc^2.
3)a and b are real positive numbers. a^n=a+1 and b^2n=b+3
n ∈N and n >1. Compare a with b.
1. $x+\sqrt{x^2+\frac{4}{x^2}+4}=4$

$\sqrt{x^2+\frac{4}{x^2}+4}=4-x$

$x^2+\frac{4}{x^2}+4=(4-x)^2$

$x^2+\frac{4}{x^2}+4=16-8x+x^2$

$\frac{4}{x^2}+4=16$

$4+4x^2=16x^2$

$12x^2-4=0$

$3x^2-1=0$

$x=\pm\sqrt{\frac{1}{3}}$

3. Originally Posted by ecMathGeek
1. $x^2+\frac{4}{x^2}+4=16-8x+x^2$

$\frac{4}{x^2}+4=16$
This step should be
$\frac{4}{x^2}+4=16-8x$

I was going to solve it using Newton's method, but when trying out different values for $x_o$ I ended up trying x=-.5

Turns out that was the answer

4. Originally Posted by ecMathGeek
1. $x+\sqrt{x^2+\frac{4}{x^2}+4}=4$

$\sqrt{x^2+\frac{4}{x^2}+4}=4-x$

$x^2+\frac{4}{x^2}+4=(4-x)^2$

$x^2+\frac{4}{x^2}+4=16-8x+x^2$

$\frac{4}{x^2}+4=16$

$4+4x^2=16x^2$

$12x^2-4=0$

$3x^2-1=0$

$x=\pm\sqrt{\frac{1}{3}}$
Ok, but when I replace √(1/3) instead of x we see that the result is not 4.

5. Originally Posted by blertta
2)a, b and c are the sides of a triangle and p , q are positive numbers and p+q = 1 . Prove that pa^2+qb^2≥pqc^2.
Assume
$pa^2+qb^2 < pqc^2$

Let c be the hypotenuse, since it is the longest side of a triangle, and therefore the greatest number.
Then by Pythagoreans theorem, $a^2+b^2=c^2$

We can substitute $a^2+b^2=c^2$ into our equation:
$pa^2+qb^2 < pq(a^2+b^2)$

And simplify:
$pa^2+qb^2 < pqa^2+pqb^2$

Now we know that p and q are between zero and one, thus pq < p and pq < q
Which means that:
$pqa^2
$pqb^2

So we can say that there are values $v_1 \mbox{ and } v_2$ which are greater than zero, where:
$pqa^2+v_1=pa^2$
$pqb^2+v_2=qb^2$

Now substituting these values into our equation:
$pqa^2+v_1+pqb^2+v_2 = pqa^2+pqb^2$

And subtracting $pqa^2 + pqb^2$ from both sides we get
$v_1+v_2 = 0$

Since $v_1 \mbox{ and } v_2$ are greater than zero, their sum is greater than zero:
$v_1+v_2 > 0$
but this contradicts that they equal zero.
$v_1+v_2 = 0$

So our assumption that $pqc^2$ is greater than $pa^2 + qb^2$ must be incorrect.
Therefore $c^2$ is not greater than $pa^2 + qb^2.$
So $c^2$ must be either less than or equal to $pa^2 + qb^2$

6. I graphed the first one, and found that x=1 is also a solution, I don't know how to find it mathematically, though, except with Newton's method.

edit:
Okay, I got it.

$x+\sqrt{x^2+\frac{4}{x^2}+4}=4$

$\sqrt{x^2+\frac{4}{x^2}+4}=4-x$

$x^2+\frac{4}{x^2}+4=16-8x+x^2$

$8x + \frac{4}{x^2}+-12=0$

$2x + \frac 1{x^2}-3=0$

$2x^3 -3x^2 +1=0$

Now there is a formula, I think it was Gauss who found it, but it's been a few semesters so I can't remember it exactly. It says something along the lines of:
If you have a polynomial, and p is one of the factors of the coefficient of the first term (so I guess that would be $p\in \{1, 2, -1, -2\}$), and q is a factor of the coefficient of the second term (so $q\in \{1, 3, -1, -3\}$) then if there is a factor, it will take the form of $(x+\frac pq) \mbox{ or } (x+ \frac qp)$

In this case, lets use q/p, where q=1 and p=2

then I did long division (I forgot how to do short division >.<)

$\begin{array}{rr} &\begin{array}{r} 2x^2-4x+2 \end{array} \\ \begin{array}{r} \\(x+\frac 12) \end{array} & \begin{array}{|r} \hline\\ 2x^3-3x^2+0x+1 \end{array} \end{array}$

So $(x+1/2)$ is a factor, giving us
$(x+\frac 12)(2x^2 -4x +2)=0$

$\frac 12(x+\frac 12)(x^2 -2x +1)=0$

$(x+\frac 12)(x -1)^2=0$

So x= -.5 or x=1

7. Hello, blertta!

1) Solve: . $x+ \sqrt{x^2+\frac{4}{x^2} +4}\:=\:4$

We have: . $\sqrt{x^2 + \frac{4}{x^2} + 4} \;=\;4 - x$

Square both sides: . $x^2 + \frac{4}{x^2} + 4 \;=\;16 - 8x + x^2 \quad\Rightarrow\quad 8x - 12 + \frac{4}{x^2} \;=\;0
$

Multiply by $\frac{x^2}{4}\!:\;\;2x^3 - 3x^2 + 1 \;=\;0$

We find that $x = 1$ is a root.

The cubic factors: . $(x-1)^2(2x+1) \;=\;0$

Answers: . $\boxed{x \;=\;1,\:-\frac{1}{2}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Did anyone notice that the radical contains a perfect square?

We have: . $x + \sqrt{x^2 + 4 + \frac{4}{x^2}} \;=\;4\quad\Rightarrow\quad x + \sqrt{\left(x+\frac{2}{x}\right)^2} \;=\;4$

Then: . $x \pm \left(x + \frac{2}{x}\right) \;=\;4$ . . . which yields the two roots above.

8. Originally Posted by Soroban
Did anyone notice that the radical contains a perfect square?

We have: . $x + \sqrt{x^2 + 4 + \frac{4}{x^2}} \;=\;4\quad\Rightarrow\quad x + \sqrt{\left(x+\frac{2}{x}\right)^2} \;=\;4$

Then: . $x \pm \left(x + \frac{2}{x}\right) \;=\;4$ . . . which yields the two roots above.

[/size]
I noticed that the terms were squares, but I didn't see they were a perfect square, I don't have enough exposure to unusual formats >.<

Good spot, though.

9. Originally Posted by angel.white
I noticed that the terms were squares, but I didn't see they were a perfect square, I don't have enough exposure to unusual formats >.<

Good spot, though.
Ok,thanks but what's about with other exercises???

10. Originally Posted by Soroban
Hello, blertta!

We have: . $\sqrt{x^2 + \frac{4}{x^2} + 4} \;=\;4 - x$

Square both sides: . $x^2 + \frac{4}{x^2} + 4 \;=\;16 - 8x + x^2 \quad\Rightarrow\quad 8x - 12 + \frac{4}{x^2} \;=\;0
$

Multiply by $\frac{x^2}{4}\!:\;\;2x^3 - 3x^2 + 1 \;=\;0$

We find that $x = 1$ is a root.

The cubic factors: . $(x-1)^2(2x+1) \;=\;0$

Answers: . $\boxed{x \;=\;1,\:-\frac{1}{2}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Did anyone notice that the radical contains a perfect square?

We have: . $x + \sqrt{x^2 + 4 + \frac{4}{x^2}} \;=\;4\quad\Rightarrow\quad x + \sqrt{\left(x+\frac{2}{x}\right)^2} \;=\;4$

Then: . $x \pm \left(x + \frac{2}{x}\right) \;=\;4$ . . . which yields the two roots above.

Thank you but what's about other exercises?

11. Originally Posted by blertta
Thank you but what's about other exercises?
I got the 2nd one earlier.

For the third, b < a

I just solved each for n, then graphed both functions, the blue is the graph for a(n), the teal is the graph for b(n).

12. Originally Posted by angel.white
This step should be
$\frac{4}{x^2}+4=16-8x$

I was going to solve it using Newton's method, but when trying out different values for $x_o$ I ended up trying x=-.5

Turns out that was the answer
Opps, haha... ty for pointing that out. It was a foolish oversight on my part.

13. Originally Posted by angel.white
then I did long division (I forgot how to do short division >.<)
for division of polynomials the short version would be synthetic division. i don't know how to do that either. i'm sure it's easy, but never had any interest in learning it, i'm good and fast enough at long division of polynomials.

$\begin{array}{rr} &\begin{array}{r} 2x^2-4x+2 \end{array} \\ \begin{array}{r} \\(x+\frac 12) \end{array} & \begin{array}{|r} \hline\\ 2x^3-3x^2+0x+1 \end{array} \end{array}$
nice use of LaTeX here!

(i wonder if i can improve on the code, maybe not. but there must be a way of not using so many arrays)

14. Hello, angel.white!

I noticed that the terms were squares, but I didn't see they were a perfect square,
I don't have enough exposure to unusual formats. . . . . Who does?
Well, okay, I do!

To tell the truth, I learned about these interesting forms during Calculus
. . while working with Arc Length problems.

The arc length formula is: . $L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$

We must take the derivative of the function, square it, add one, and take the square root.
. . Then we must integrate this ugly function.

If the problem's author is humane, the expression simplifies very nicely.
However, by necessity, the original function is usually strange-looking.

Here's an example: . $y \:=\:\frac{1}{6}x^3 + \frac{1}{2x} \;=\;\frac{1}{6}x^3 + \frac{1}{2}x^{-1}$

(Seriously, who would want to graph that curve anyway
. . and find the length of a portion of it?)

We have: . $\frac{dy}{dx} \:=\:\frac{1}{2}x^2 - \frac{1}{2}x^{-2} \:=\;\frac{1}{2}\left(x^2-\frac{1}{x^2}\right)$

Then: . $\left(\frac{dy}{dx}\right)^2 \;=\;\frac{1}{4}\left(x^2-\frac{1}{x^2}\right)^2 \;=\;\frac{1}{4}\left({\color{red}x^4 - 2 + \frac{1}{x^4}}\right)$ .[1]

And: . $1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \frac{1}{4}\left(x^4 - 2 + \frac{1}{x^4}\right)$

. . $= \;\frac{1}{4}\left(x^4 - 2 + \frac{1}{x^4} + 4\right) \;= \;\frac{1}{4}\left({\color{red}x^4 + 2 + \frac{1}{x^4}}\right)$

. . which looks like [1] except for a sign-change.
Hence, we are expected to recognize it as: . $\frac{1}{4}\left(x^2 + \frac{1}{x^2}\right)^2$

Finally: . $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\frac{1}{4}\left(x^2 + \frac{1}{x^2}\right)^2} \;=\;\frac{1}{2}\left(x^2 + \frac{1}{x^2}\right)$

And now we can integrate: . $L \;=\;\frac{1}{2}\int^b_a\left(x^2 + x^{-2}\right)\,dx$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you haven't had Calculus yet, it's still an interesting bit of algebra.

It's a great reputation-maker.

Glance at $x^8 + 6 + \frac{9}{x^8}$ and say: .Obviously, it's $\left(x^4 + \frac{3}{x^4}\right)^2$
. . and ride off into the sunset.

15. Blertta, I don't know if you saw this, but I got problem #2 in post 5: http://www.mathhelpforum.com/math-he...513-post5.html
Originally Posted by Jhevon
nice use of LaTeX here!
Thank you, it's my first use of arrays ^_^