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Thread: Radicals

  1. #1
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    Radicals

    I have no clue how to do these, and the book I have does not help at all.
    Can someone help me please?

    $\displaystyle (\sqrt{3} + \sqrt{5})^2$
    On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15}$


    Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5})$

    Solve.$\displaystyle \sqrt[3] {x-7}=2$

    Solve$\displaystyle \sqrt{2x+1}=2x-5$
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  2. #2
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    Quote Originally Posted by alexl13 View Post
    I have no clue how to do these, and the book I have does not help at all.
    Can someone help me please?

    $\displaystyle (\sqrt{3} + \sqrt{5})^2$
    On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15}$ e^(i*pi): you need only combine those radicals like you would with rational numbers and factorise


    Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5})$

    Solve.$\displaystyle \sqrt[3] {x-7}=2$

    Solve$\displaystyle \sqrt{2x+1}=2x-5$
    1. Use FOIL as normal:

    $\displaystyle (\sqrt{3} + \sqrt{5})^2 = 3 + 2\sqrt{3}\sqrt{5} + 5 = 8 + 2\sqrt{15} = 2(4+\sqrt{15})$

    2. It's the same as above but with different numbers

    3. Cube to remove the radical:

    $\displaystyle x-7 = 2^3 = 8$ - easy to see that x=15

    4. Square to remove the radical, use FOIL on the RHS and solve the resulting quadratic
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  3. #3
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    Quote Originally Posted by alexl13 View Post
    I have no clue how to do these, and the book I have does not help at all.
    Can someone help me please?

    $\displaystyle (\sqrt{3} + \sqrt{5})^2$
    On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15} \textcolor{red}{= 8 + 2\sqrt{15}}$ ... that's all.


    Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5}) \textcolor{red}{= 19 + 2\sqrt{90} = 19 + 3\sqrt{10}}$



    Solve.$\displaystyle \sqrt[3] {x-7}=2$

    cube both sides ...

    $\displaystyle \textcolor{red}{x-7 = 8}$

    $\displaystyle \textcolor{red}{x = 15}$


    Solve$\displaystyle \sqrt{2x+1}=2x-5$

    square both sides, then solve the resulting quadratic ... don't forget to check your results in the original equation. go ahead, try it.
    .
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  4. #4
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    Quote Originally Posted by alexl13 View Post
    I have no clue how to do these, and the book I have does not help at all.
    Can someone help me please?

    $\displaystyle (\sqrt{3} + \sqrt{5})^2$
    On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15}$


    Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5})$

    Solve.$\displaystyle \sqrt[3] {x-7}=2$

    Solve$\displaystyle \sqrt{2x+1}=2x-5$
    FOIL

    $\displaystyle (\sqrt{3}+\sqrt{5})^2=(\sqrt{3})^2+\overbrace{\sqr t{3}\sqrt{5}+\sqrt{3}\sqrt{5}}^{\text{like terms}}+(\sqrt{5})^2$

    $\displaystyle =3+2\sqrt{3}\sqrt{5}+5=3+2\sqrt{15}+5=8+2\sqrt{15}$
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by skeeter View Post
    $\displaystyle
    (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5}) \textcolor{red}{= 19 + 2\sqrt{90} = 19 + 3\sqrt{10}}
    $
    Where did you get get 90 from in Q2? I got $\displaystyle 19+\sqrt{70}$ with $\displaystyle \sqrt{70} = \sqrt{14}\sqrt{5}$
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  6. #6
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    Quote Originally Posted by e^(i*pi) View Post

    4. Square to remove the radical, use FOIL on the RHS and solve the resulting quadratic
    I got $\displaystyle 2x+1=4x^2-20x=25$
    but what confuses me is the exponent.
    Would I make is a square root so that it would be 4x?
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    Where did you get get 90 from in Q2? I got $\displaystyle 19+\sqrt{70}$ with $\displaystyle \sqrt{70} = \sqrt{14}\sqrt{5}$
    my mistake ... i was thinking 5 times 18 for some reason.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by alexl13 View Post
    I got $\displaystyle 2x+1=4x^2-20x=25$
    but what confuses me is the exponent.
    Would I make is a square root so that it would be 4x?
    $\displaystyle (\sqrt{2x+1})^2=(2x-5)^2$

    $\displaystyle 2x+1=4x^2-10x+25$


    $\displaystyle 4x^2-12x+24=0$

    $\displaystyle x^2-3x+6=0$

    $\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(6)}}{2(1)}$
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  9. #9
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    Quote Originally Posted by alexl13 View Post
    I got $\displaystyle 2x+1=4x^2-20x=25$
    but what confuses me is the exponent.
    Would I make is a square root so that it would be 4x?
    No you can solve the quadratic using factorising, completing the square or the quadratic equation


    $\displaystyle
    (\sqrt{2x+1})^2=(2x-5)^2
    $

    $\displaystyle 2x+1 = 4x^2 - 20x + 25$

    $\displaystyle 4x^2 - 22x +25 = 0$

    $\displaystyle x = \frac{22 \pm \sqrt{(-22)^2 - 4(4)(25)}}{2(4)}$

    As $\displaystyle b^2-4ac > 0$ there will be two, distinct real solutions
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