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    Where is the error in this ...

    (x+5)^(1/2)=x+3

    x+5=x^2+6x+9

    x^2+5x+4=0

    (x+4)(x+1)=0

    hence x=-4 or x=-1
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    (Original post by Econowizard)
    Where is the error in this ...

    (x+5)^(1/2)=x+3

    x+5=x^2+6x+9

    x^2+5x+4=0

    (x+4)(x+1)=0

    hence x=-4 or x=-1
    Check that both solutions work in the original equation. When you square both sides of an equation, you will often generate extra solutions. E.g.

    x=1 has one solution

    x^2 =1 has two solutions.

    For most algebraic manipulation from step 1 to step 2, step 1 implies step 2 and step 2 implies step 1. For example, adding 2 to both sides of an equation.

    But in cases such as squaring, this isn't true so the final solutions that you get many not satisfy the original equation.
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    if you sketch the graphs of y = ( x + 5 ).5 and y = x + 3 you can see they only intersect once.
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    (Original post by Econowizard)
    Where is the error in this ...

    (x+5)^(1/2)=x+3

    x+5=x^2+6x+9

    x^2+5x+4=0

    (x+4)(x+1)=0

    hence x=-4 or x=-1
    I'd solve this slightly differently:

    Let y=x+5

    y^\frac{1}{2}=y-2

    \therefore y-y^\frac{1}{2}-2=0

    \therefore (y^\frac{1}{2}+1)(y^\frac{1}{2}-2) =0

    \therefore y^\frac{1}{2}=-1, 2

    \therefore y=4 \implies x=-1 (for real solutions)
 
 
 
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