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  1. #1
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    help me solve this square root equation

    sqrt(10-3x) + sqrt(x^2+5) = 7
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  2. #2
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    Quote Originally Posted by cityismine View Post
    sqrt(10-3x) + sqrt(x^2+5) = 7
    $\displaystyle \sqrt{10-x}+\sqrt{x^2+5}=7$

    The domain is $\displaystyle x \in (-\infty~,~10]$

    The general procedure with such a type of equation is:

    Square both sides - isolate remaining roots at one side - square again - isolate ... and so on:

    $\displaystyle 10-x + 2\sqrt{10-x} \cdot \sqrt{x^2+5}+x^2+5 = 49~\implies~ 2\sqrt{-x^3+10x^2-5x+50}=-x^2+x+34$ ...... Square again:

    $\displaystyle 4(-x^3+10x^2-5x+50)=x^4-2x^3-67x^2+68x+1156$ ...... Collect like terms:

    $\displaystyle x^4+2x^3-107x^2+88x+956 = 0$

    I don't know how to solve such an equation algebraically. So I used my computer which calculated 4 approximate solutions:
    $\displaystyle \left| \begin{array}{l}x_1\approx -11.4101 \\ x_2 \approx -2.6234 \\ x_3 \approx 3.95199 \\ x_4 \approx 8.08147\end{array} \right.$

    Since squaring is not an equivalent operation you must check these values by plugging them into the original equation and check if they satisfy the equation.
    Better use a calculator
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  3. #3
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    Hello,

    Quote Originally Posted by earboth View Post
    $\displaystyle \sqrt{10-{\color{red}3}x}+\sqrt{x^2+5}=7$

    [...]
    You missed a 3...

    $\displaystyle \sqrt{10-3x}+\sqrt{x^2+5}=7$

    Domain is $\displaystyle \left]-\infty~,~\frac{10}{3}\right]$

    Squaring :

    $\displaystyle (10-3x)+(x^2+5)+2 \cdot \sqrt{(10-3x)(x^2+5)}=49$

    $\displaystyle 2 \cdot \sqrt{(10-3x)(x^2+5)}=-x^2+3x+34$

    Squaring :

    $\displaystyle 4 (10-3x)(x^2+5)=x^4+9x^2+34^2-6x^3-68x^2+204x$

    $\displaystyle 40x^2+200-12x^3-60x=x^4-6x^3-59x^2+204x+1156$

    $\displaystyle 0=x^4+6x^3-99x^2+264x+956$

    -2 is a solution

    $\displaystyle 0=(x+2)(x^3+4x^2-107x+478)$

    And it seems that the remaining has no real solution.
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