Algebra help

May 2010
8
0
How do you solve equations with radicals in them like this

2√x-3=5

Any help would be great.
 

Prove It

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Aug 2008
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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

\(\displaystyle 2\sqrt{x} - 3 = 5\)?
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
\(\displaystyle 2\sqrt{x - 3} = 5\)

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

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

\(\displaystyle x - \frac{12}{4} = \frac{25}{4}\)

\(\displaystyle x = \frac{37}{4}\).
 
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May 2010
8
0

skeeter

MHF Helper
Jun 2008
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May 2010
8
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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.
 
Jun 2009
806
275
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.
 
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May 2010
8
0
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.
 
Last edited:
Jun 2009
806
275
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

\(\displaystyle x = 2\sqrt(ab)\)
 
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