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Math Help - Help on Radical Equations

  1. #1
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    Help on Radical Equations

    Please help me on these equations. Please click the attachment
    Attached Files Attached Files
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  2. #2
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    Quote Originally Posted by quebec567 View Post
    Please help me on these equations. Please click the attachment
    You need to go through these adding brackets to make the meaning clear, or
    use equation editor to make the root signs extend over what you want the
    root of.

    1. \root{4}\of{x-3}=2

    raise both sides of this to the fourth power:

    {x-3}=2^4=16,

    so:

    x=16+3=19

    RonL
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  3. #3
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    Quote Originally Posted by quebec567 View Post
    Please help me on these equations. Please click the attachment
    Hello,

    2) the same method as CaptainBlack has demonstrated:

    \sqrt[4]{y+3}=3 will become:

    y+3=81 \Longleftrightarrow y = 78

    3) and 4) are ambiguously written, so I can't help you.

    5) 5=\frac{1}{\sqrt{x}} \Longleftrightarrow \sqrt{x}=\frac{1}{5} \Longrightarrow x=\frac{1}{25}

    6) \frac{2}{\sqrt{x}}=5 \Longleftrightarrow \frac{2}{5}=\sqrt{x} \Longrightarrow x=\frac{4}{25}

    7) \sqrt{3x+1}=\sqrt{2x+6} \Longrightarrow 3x+1=2x+6 \Longrightarrow x = 5.
    You have to check this answer by re-substituting the result into the original equation because squaring an equation could change something false into a true equation.

    8) and 9) are similar to 7). I leave them for you.

    10) \sqrt{x^2+6}+x-3=0 Isolate the square-root at the LHS of the equation:

    \sqrt{x^2+6}=-x+3 \Longrightarrow x^2+6=x^2-6x+9 \Longrightarrow -3=-6x \Longrightarrow x=\frac{1}{2}

    EB
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  4. #4
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    Hello, quebec567!

    Captain Black is absolutely right.
    . . Unless you use parentheses, we have to guess what you meant.

    Here are a few more . . .


    #2 is the same as #1
    3)\;\;\sqrt[3]{6x + 9} + 9 \:=\:6
    Subtract 9 from both sides: . \sqrt[3]{6x + 9}\:=\:-3

    Cube both sides: . \left(\sqrt[3]{6x + 9}\right)^3\;=\;(-3)^3

    We have: . 6x + 9 \:=\:-27\quad\Rightarrow\quad 6x\:=\:-36\quad\Rightarrow\quad x \,=\,-6


    #4 is the same as #3.
    5)\;\;5 \:=\:\frac{1}{\sqrt{x}}
    We have: . \sqrt{x}\:=\:\frac{1}{5}

    Square both sides: . \left(\sqrt{x}\right)^2\:=\:\left(\frac{1}{5}\righ  t)^2\quad\Rightarrow\quad x\,=\,\frac{1}{25}


    #6 is the same as #5.
    8)\;\;\sqrt{5x+3} - \sqrt{2x+3}\:=\:0
    We have: . \sqrt{5x+3} \:=\:\sqrt{2x+3}

    Square both sides: . \left(\sqrt{5x+3}\right)^2\:=\:\left(\sqrt{2x+3}\r  ight)^2\quad\Rightarrow\quad5x + 3\:=\:2x+3

    Therefore: . 3x\,=\,0\quad\Rightarrow\quad x\,=\,0


    12)\;\;\sqrt{x-6}\:=\:\sqrt{3} - \sqrt{x}
    Square both sides: . \left(\sqrt{x-6}\right)^2\:=\:\left(\sqrt{3} - \sqrt{x}\right)^2

    and we have: . x - 6 \:=\:3 - 2\sqrt{3x} + x\quad\Rightarrow\quad 2\sqrt{3x}\:=\:9\quad\Rightarrow\quad \sqrt{3x}\:=\:\frac{9}{2}

    Square both sides: . \left(\sqrt{3x}\right)^2\:=\:\left(\frac{9}{2}\rig  ht)^2\quad\Rightarrow\quad3x\:=\:\frac{81}{4}

    Divide by 3: . x\:=\:\frac{27}{4}

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  5. #5
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    What if the equation contains a principal root? like Number 9:
    9.) 3√2x+3 = √y+10
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  6. #6
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    Quote Originally Posted by quebec567 View Post
    What if the equation contains a principal root? like Number 9:
    9.) 3√2x+3 = √y+10
    Hello,

    1. I presume that there must be a typo because you can't solve an equation with 2 variables.

    2. Let's pretend that the equation is:

    3\sqrt{2x+3}=\sqrt{x+10}. Now square both sides of the equation:

    9(2x+3)=x+10 \Longrightarrow 18x+27=x+10 \Longrightarrow 17x = -17 \Longrightarrow x = -1

    Now plug in this solution into the original equation and check it.

    EB

    Please do us a favour and use brackets. For instance your equation is clearly readable if you had written:

    9.) 3√(2x+3) = √(y+10)
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