# Algebra help

• May 13th 2010, 04:48 PM
cw86
Algebra 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 PM
Prove It
Quote:

Originally Posted by cw86
How do you solve equations with radicals in them like this

2√x-3=5

Any help would be great.

Is this

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

or

$\displaystyle 2\sqrt{x} - 3 = 5$?
• May 13th 2010, 04:55 PM
cw86
Quote:

Originally Posted by Prove It
Is this

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

or

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

I'm sorry its the first one square root
• May 13th 2010, 04:57 PM
Prove 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 PM
cw86
Quote:

Originally Posted by Prove It
$\displaystyle 2\sqrt{x - 3} = 5$

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

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

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

$\displaystyle x = \frac{37}{4}$.

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 PM
skeeter
Quote:

Originally Posted by cw86
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.

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 PM
cw86
Quote:

Originally Posted by skeeter
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.

That wont work cause I don't know how long a is.
• May 13th 2010, 06:11 PM
sa-ri-ga-ma
Quote:

Originally Posted by cw86
That wont work cause I don't know how long a is.

No need to know a. According to Pythagoras

$\displaystyle (a+b)^2 = x^2 + (a - b)^2$

Simplify the above equation to get the result.
• May 13th 2010, 07:38 PM
cw86
Quote:

Originally Posted by sa-ri-ga-ma
No need to know a. According to Pythagoras

$\displaystyle (a+b)^2 = x^2 + (a - b)^2$

Simplify the above equation to get the result.

Lets say Radius b is 2 and line x is 5 what would the radius of a be?,I'm not understanding how you can solve it with the Pythagorean Theorem.
• May 13th 2010, 07:50 PM
sa-ri-ga-ma
Quote:

Originally Posted by cw86
Lets say Radius b is 2 and line x is 5 what would the radius of a be?,I'm not understanding how you can solve it with the Pythagorean Theorem.

In the problem no numerical values are included. You have to prove

$\displaystyle x = 2\sqrt(ab)$
• May 13th 2010, 08:00 PM
cw86
Quote:

Originally Posted by sa-ri-ga-ma
In the problem no numerical values are included. You have to prove

$\displaystyle x = 2\sqrt(ab)$

• May 13th 2010, 08:04 PM
Bacterius
(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 PM
sa-ri-ga-ma
Quote:

Originally Posted by cw86
Have you simplified $\displaystyle (a+b)^2 - (a-b)^2 = x^2$?