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  1. #1
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    Algebra help

    How do you solve equations with radicals in them like this

    2√x-3=5

    Any help would be great.
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  2. #2
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    Quote Originally Posted by cw86 View Post
    How do you solve equations with radicals in them like this

    2√x-3=5

    Any help would be great.
    Is this

    2\sqrt{x - 3} = 5

    or

    2\sqrt{x} - 3 = 5?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Is this

    2\sqrt{x - 3} = 5

    or

    2\sqrt{x} - 3 = 5?
    I'm sorry its the first one square root
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  4. #4
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    2\sqrt{x - 3} = 5

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

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

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

    x = \frac{37}{4}.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    2\sqrt{x - 3} = 5

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

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

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

    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.
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  6. #6
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    Quote Originally Posted by cw86 View Post
    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.



    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.
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  7. #7
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    Quote Originally Posted by skeeter View Post



    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.
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  8. #8
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    Quote Originally Posted by cw86 View Post
    That wont work cause I don't know how long a is.
    No need to know a. According to Pythagoras

    (a+b)^2 = x^2 + (a - b)^2

    Simplify the above equation to get the result.
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  9. #9
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    Quote Originally Posted by sa-ri-ga-ma View Post
    No need to know a. According to Pythagoras

    (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 by cw86; May 13th 2010 at 07:52 PM.
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  10. #10
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    Quote Originally Posted by cw86 View Post
    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

    x = 2\sqrt(ab)
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  11. #11
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    Quote Originally Posted by sa-ri-ga-ma View Post
    In the problem no numerical values are included. You have to prove

    x = 2\sqrt(ab)
    This is becoming painful
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  12. #12
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    (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 ?
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  13. #13
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    Quote Originally Posted by cw86 View Post
    This is becoming painful
    Why? What is required in the problem?

    Have you simplified (a+b)^2 - (a-b)^2 = x^2?

    Do it first. Then let us think about your pain.
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  14. #14
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    Thanks for all the help guys.
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