How do you solve equations with radicals in them like this

2√x-3=5

Any help would be great.

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- May 13th 2010, 04:48 PMcw86Algebra help
How do you solve equations with radicals in them like this

2√x-3=5

Any help would be great. - May 13th 2010, 04:53 PMProve It
- May 13th 2010, 04:55 PMcw86
- May 13th 2010, 04:57 PMProve It
$\displaystyle 2\sqrt{x - 3} = 5$

$\displaystyle \sqrt{x - 3} = \frac{5}{2}$

$\displaystyle x - 3 = \left(\frac{5}{2}\right)^2$

$\displaystyle x - \frac{12}{4} = \frac{25}{4}$

$\displaystyle x = \frac{37}{4}$. - May 13th 2010, 05:08 PMcw86
How would you solve for radius a with only knowing radius b and the length of x?

at this website Tangent circles, Common external tangent line, Geometric Mean. Elearning. - May 13th 2010, 05:40 PMskeeter
http://www.gogeometry.com/problem/p3...nt_circles.gif

draw a segment parallel to CD from point B to AC ... let the intersection point with AC be point E.

right triangle ABE ... vertical leg = a-b , horizontal leg = x , hypotenuse = a+b

now use Pythagoras to get the desired result.

... btw, next time start a new problem w/ a new thread. - May 13th 2010, 05:57 PMcw86
- May 13th 2010, 06:11 PMsa-ri-ga-ma
- May 13th 2010, 07:38 PMcw86
- May 13th 2010, 07:50 PMsa-ri-ga-ma
- May 13th 2010, 08:00 PMcw86
- May 13th 2010, 08:04 PMBacterius
(To Prove It) If you read my original message, sorry, didn't read properly, must be tired xD

cw86, the Pythagorean Theorem allows you to recover the length of one side of a right angled triangle provided the two other sides. Now, plugging in the values of "a" and "x" in the equation and solving for "b", you will have your answer :)

Does it make sense ? Or do you need further help ? - May 13th 2010, 08:12 PMsa-ri-ga-ma
- May 13th 2010, 11:05 PMcw86
Thanks for all the help guys.