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kristo · Apr 16, 2007, 07:53 AM

Hey, I need some help with solving x for these equations:
1 $x^2-4x+6=sqrt{2x^2-8x+12}$

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

3 $\frac{sqrt x + \sqrt[3]x}{sqrt x - \sqrt[3]x}=3$

4 $sqrt{2x^2+21x-11}-sqrt{2x^2-9x+4}=sqrt{18x-9}$

5 $sqrt{3x^2+6x+7}+sqrt{5x^2+10x+14}=4-2x-x^2 \\$

And how do I express X in these equations?

6 $\frac{x}{x+a} + \frac {sqrt{a}}{sqrt{x+a}}=\frac{b}{a}$

7 $\frac{sqrt{a+x}+ sqrt{a-x}}{sqrt{a+x}-sqrt{a-x}}=sqrt{b}$

Thanks in advance :)

//My attempt at solving the first one:
Let $x^2-4x+6$ be a, so I get an equation:
$a=sqrt{2a}$ squaring both sides it will be
$a^2=2a$
And if I replaced the As I would get a difficult equation with 4th powers, isn't there an easier way to solve it?

I have the same problem with the 2nd one, it goes to high powers and then I'm not able to solve it.

galactus · Apr 16, 2007, 09:47 AM

Hey, I need some help with solving x for these equations:
1 $x^2-4x+6=sqrt{2x^2-8x+12}$

I will step thorugh this one. Maybe it'll help you conquer the others. Okey-doke?

Factor is the ticket. Factor what's inside the radical.

$x^{2}-4x+6=\sqrt{2(x^{2}-4x+6)}$

See something coming together?

Divide through by $\sqrt{x^{2}-4x+6}$ :

$\sqrt{x^{2}-4x+6}=\sqrt{2}$

Square both sides:

$x^{2}-4x+6=2$

$x^{2}-4x+4$

$(x-2)^{2}=0$

$x=2$

BTW, the second one has no solution. Unless there's a typo.

kristo · Apr 17, 2007, 06:48 AM

Thanks a lot Galactus, I've solved some of the equations now, still need help with others.

$\frac{sqrt{a+x}+ sqrt{a-x}}{sqrt{a+x}-sqrt{a-x}}=sqrt{b}$
First I divide both sides by $sqrt{a+x} - sqrt{a-x} \\$
Then I get $sqrt{a+x}+sqrt{a-x}= sqrt{b}(sqrt{a+x}-sqrt{a+x}) \\$
and dividing it by $sqrt{b}$ I get
$\frac{sqrt{a+x}+sqrt{a-x}}{sqrt{b}}=sqrt{a+x}-sqrt{a-x} \\$
Then i multiply both sides by $sqrt{a+x}+sqrt{a-x}$
$\frac1{sqrt{b}}=(sqrt{a+x}+sqrt{a-x})(sqrt{a+x}-sqrt{a-x}) \\$
$\frac1{sqrt{b}}=a+x-a+x \\$
$2x=\frac{1}{sqrt{b}} \\$
$x=\frac{1}{2sqrt{b}} \\$
Removing the irrationality it will be
$x=\frac{sqrt{b}}{2b}$

I also solved the 4th one:
$sqrt{2x^2+21x-11}-sqrt{2x^2-9x+4}=sqrt{18x-9}$
Factoring everything I get
$sqrt{(2x-1)(x+11)} - sqrt{(2x-1)(x-4)}=3sqrt{2x-1} \\$
so I can divide everything by $sqrt{2x-1}$
$sqrt{x+11}-sqrt{x-4}=3$
Too lazy to write the rest in latex lol, x should solve for 5.

Any hints for 3rd, 5th and 6th?

Capuchin · Apr 17, 2007, 06:50 AM

You seem to be mixing up multiplication and division in the first one you have done, kristo. Your 6th and 7th line don't follow. And on the 2nd line you say you divide, but you actually multiply.

On the 6th line you say you multiply both sides, but you actually divide one side and multiply the other!! Not good!!

kristo · Apr 17, 2007, 07:02 AM

Originally Posted by Capuchin

You seem to be mixing up multiplication and division in the first one you have done, kristo. Your 6th and 7th line dont follow. And on the 2nd line you say you divide, but you actually multiply.

On the 6th line you say you multiply both sides, but you actually divide one side and multiply the other!!!! Not good!!!

You're fast lol, I just noticed the mistake and was going to edit my post
I'll try to do it again, going to post later if I get the solution.

kristo · Apr 17, 2007, 07:39 AM

I think I got it now.

$\frac{sqrt{a+x}+ sqrt{a-x}}{sqrt{a+x}-sqrt{a-x}}=sqrt{b}$
First I multiply both sides by $sqrt{a+x} - sqrt{a-x} \\$
Then I get $sqrt{a+x}+sqrt{a-x}= sqrt{b}(sqrt{a+x}-sqrt{a+x}) \\$

Here I multiply both sides by $sqrt{a+x}-sqrt{a-x}$ and get
$a+x-a+x=sqrt{b}(sqrt{a+x}-sqrt{a-x})^2$
$2x=sqrt{b}(a+x-2(a+x-a+x)+a-x)$
$2x=sqrt{b}(2a-4x)$
now I can divide both sides by 2
$x=a sqrt{b}-2x sqrt{b}$
$x+2x sqrt{b}=a sqrt{b}$
$x=\frac{a sqrt{b}}{1+2 sqrt{b}}$
Removing the irrationality
$x= \frac{a sqrt{b}-2ab}{a-4b}$

Capuchin · Apr 17, 2007, 07:42 AM

That looks much better :), at a glance

galactus · Apr 17, 2007, 09:23 AM

3 $\frac{sqrt x + \sqrt[3]x}{sqrt x - \sqrt[3]x}=3$

Here's one of many ways:

Rewrite this one as $x^{\frac{1}{2}}+x^{\frac{1}{3}}=3x^{\frac{1}{2}}-3x^{\frac{1}{3}}$

$-2x^{\frac{1}{2}}+4x^{\frac{1}{3}}=0$

Factor out x^(1/3):

$x^{\frac{1}{3}}(-2x^{\frac{1}{6}}+4)$

OK, what makes $-2x^{\frac{1}{6}}+4=0$

In other words, solve $x^{\frac{1}{6}}=2$