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Math Help - help me out this root problems

  1. #1
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    Question help me out this root problems

    question:1. √[3(√x-1/√x)]=2, then √[x]+1/√[x]= what....

    question 2 If a=2b and b =4c, then √[a/16bc]

    question 3 the sides of a triangle are denoted by x,y,z. Area of the triangle and semi-perimeter of the triangle are denoted by P and q respectively. If P =√[q(q-x)(q-y)(q-z) and x+y-z = y+z-x = z+x-y = 4. Find P [in square units]

    question 4. If a is any natural number, then a-1/a will always be greater than or equal to
    option 1. 3{a+1/a} option 2. a+1/a option 3. [a+1/a] option 4. 3(a+1/a)

    i am only 8th... so please explain clearly.. i already posted the question.. i didn't get the answer and so i again posted.. this time help me out..please..

    thanks a lot...
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  2. #2
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    Re: help me out this root problems

    Hi Kohila,

    Here are some thoughts on Question 2: If a=2b and b =4c, then √[a/16bc].

    *It requires to determine the value of √[a/16bc]. Since the exact values of a, b, c are not provided, we need to know the relationships among a, b, c and make appropriate substitutions to factorize √[a/16bc].
    *The provided constraints, i.e., a=2b and b=4c, are just what we want to know---the relationships among a, b, c.
    *From a=2b and b=4c we observe that both a and c relate to b. So we could substitute a and c in √[a/16bc] with b. More specifically, we have a=2b and c=b/4.
    *Now transform √[a/16bc] to √[(2b)/16b(b/4)] and I'm sure you could proceed with the rest
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  3. #3
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    Re: help me out this root problems

    Hi Kohila,

    Here are some thoughts on Question 3. Hope they are helpful.

    *the key point to solve p is to determine the values of q, q-x, q-y, q-z.
    *semi-perimeter q=(x+y+z)/2, then we have:
    q-x=(y+z-x)/2 where (y+z-x)'s value is assigned. So we can determine the value of q-x and that of q-y, q-z similarly.
    *Now only the value of q=(x+y+z)/2 is left to determine. Please observe the provided constraints as follows:
    x + y - z =4
    -x + y + z =4
    x - y +z =4
    Any clue? All three variables x, y, and z happen three times in three provided equations: two times of + and one time of -. Combine all them then we get exactly (x+y+z) and therefore the value of q.

    Now you could try find the values of q, q-x, q-y, q-z accordingly~
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  4. #4
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    Re: help me out this root problems

    Hello, kohila!

    [1]\;\text{Given: }\;\sqrt[3]{3\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right)} \:=\:2

    . . . \text{Find: }\:\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}

    Cube the equation: . 3\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right) \;=\;8 \quad\Rightarrow\quad \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \;=\;\frac{8}{3}


    Square both sides:

    . . . \left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\left(\frac{8}{3}\right)^2

    . . \sqrt[3]{x^2} - 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{64}{9}

    . . . . . \sqrt[3]{x^2} + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9}


    Add 2 to both sides:

    . . \sqrt[3]{x^2} + 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9} + 2

    . . . \left(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\frac{100}{9}


    Therefore: . \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} \;=\;\frac{10}{3}
    Thanks from kohila
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  5. #5
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    Re: help me out this root problems

    Quote Originally Posted by Soroban View Post
    Hello, kohila!


    Cube the equation: . 3\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right) \;=\;8 \quad\Rightarrow\quad \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \;=\;\frac{8}{3}


    Square both sides:

    . . . \left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\left(\frac{8}{3}\right)^2

    . . \sqrt[3]{x^2} - 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{64}{9}

    . . . . . \sqrt[3]{x^2} + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9}


    Add 2 to both sides:

    . . \sqrt[3]{x^2} + 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9} + 2

    . . . \left(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\frac{100}{9}


    Therefore: . \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} \;=\;\frac{10}{3}
    wow, it's really a neat solution!
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  6. #6
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    Re: help me out this root problems

    Thanks a lot...
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