I can't figure this out for the life of me...
I just need the fastest solution please!
√x+10= 5-√3-x

2. Originally Posted by puzzledwithpolynomials
I can't figure this out for the life of me...
I just need the fastest solution please!
√x+10= 5-√3-x
Hi

First show that $\displaystyle -10 \leq x \leq 3$

Then square both sides
$\displaystyle x+10 = 25 - 10 \sqrt{3-x} + 3 - x$

Re-arrange
$\displaystyle 5 \sqrt{3-x} = 9 - x$

Square both sides again and solve the second degree polynomial equation.
Do not forget to check if the solutions are effectively inside [-10 ; 3]

3. Could you explain how you re-arranged it?

4. Originally Posted by running-gag
Hi

First show that $\displaystyle -10 \leq x \leq 3$

Then square both sides
$\displaystyle x+10 = 25 - 10 \sqrt{3-x} + 3 - x$

Re-arrange
$\displaystyle 5 \sqrt{3-x} = 9 - x$

Square both sides again and solve the second degree polynomial equation.
Do not forget to check if the solutions are effectively inside [-10 ; 3]
Originally Posted by puzzledwithpolynomials
Could you explain how you re-arranged it?
Hi puzzledwithpolynomials,

Running-gag assumed you meant this:

$\displaystyle \sqrt{x+10}=5-\sqrt{3-x}$

Is that correct, or is it:

$\displaystyle \sqrt{x}+10=5-\sqrt{3}-x$? Can't really tell.

If running-gag is correct, then the simplification would go this way:

$\displaystyle \sqrt{x+10}=5-\sqrt{3-x}$

Square both sides:

$\displaystyle x+10 = 25 - 10 \sqrt{3-x} + 3 - x$

Bring everything to the left side except for the radical term:

$\displaystyle x+10-25-3+x=-10\sqrt{3-x}$

Combine terms:

$\displaystyle 2x-18=-10\sqrt{3-x}$

Divide by -2:

$\displaystyle -x+9=5\sqrt{3-x} \ \ \ \ or \ \ \ \ 9-x=5\sqrt{3-x}$

Square both sides again to eliminate the radical:

$\displaystyle (-x+9)^2=(5\sqrt{3-x})^2$

$\displaystyle x^2-18x+81=25(3-x)$

$\displaystyle x^2-18x+81=75-25x$

Move all terms to the left, combine terms, and set the left side equal zero.

$\displaystyle x^2+7x+6=0$

$\displaystyle (x+6)(x+1)=0$

$\displaystyle x=-6 \ \ or \ \ x=-1$

Do these solutions check out if you substitute them in the original radical equation? Are they within the domain that Running-gag set up for you at the beginning?

Good luck.

5. Ahh. I see it now. Yes, running-gag was correct. I'm sorry about the notation issue.
Thanks again.