i swear i'll never understand problems with radicals >.<

anyway: $\sqrt{\sqrt{x+8}+x}=2$

show work if you can..someday i'll understand those things -.-..always confudling me..

2. Originally Posted by jcube69
i swear ill nvr understand problems with gd radicals >.<

anyways $\sqrt{\sqrt{x+8}+x}=2$

show work if you can..someday ill understand those buggers -.-..always confudling me..
square both sides, you get

$\sqrt{x + 8} + x = 4$

$\Rightarrow \sqrt{x + 8} = 4 - x$

now square both sides again and continue

(and please, do not make the mistake of thinking $(4 - x)^2 = 4^2 - x^2$, it is wrong and will tick your professor off)

3. Originally Posted by Jhevon
square both sides, you get

$\sqrt{x + 8} + x = 4$

$\Rightarrow \sqrt{x + 8} = 4 - x$

now square both sides again and continue
i did that u get $x + 8=x^2-8x+16$then $-8=x^2-9x or x^2-9x+8=0$

$\frac{9\pm\sqrt{81-32}}{2}$ $\frac{9\pm7}{2}=4.5\pm3.5$ thus $x=1,8$?

4. Originally Posted by jcube69
i did that u get $x + 8=x^2-8x+16$then $-8=x^2-9x or x^2-9x+8=0$
you're right so far

$\frac{9\pm\sqrt{81-32}}{2}$ $\frac{9\pm7}{2}=4.5\pm3.5$

thus $x=1,8$?
yes, x = 8 or x = 1 are the solutions to the quadratic. are both solutions valid for our original problem though?

also, that wasn't so hard, was it?

5. yes, x = 8 or x = 1 are the solutions to the quadratic. are both solutions valid for our original problem though?

also, that wasn't so hard, was it?
bah didnt think about that..8 woulnt work because the sqrt of 12 is not 2

thus it leaves only 1

6. Originally Posted by jcube69
bah didnt think about that..8 woulnt work because the sqrt of 12 is not 2

thus it leaves only 1
right. be aware that usually when you square things, you introduce erroneous solutions. you need to check when done

7. Originally Posted by Jhevon
right. be aware that usually when you square things, you introduce erroneous solutions. you need to check when done
yup... tyvm ^.^