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why does incorrect solution come from step 1 Watch

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    Squaring both sides of an equation like this can result in incorrect solutions.
    Consider the equation  \sqrt x=-1 , if you square both sides you get  x=1 and this is obviously not true.
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    (Original post by KloppOClock)
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    Squaring both sides of an equation can generate extra solutions. The reason is because the reverse process can have multiple solutions i.e. \pm \sqrt{x}

    In this example, -4 satisfies (x+5) = (x+3)^2

    But if you square root both sides, -4 only satisfies the equation if you take the negative square root of one of the sides:

    -\sqrt{x+5} = (x+3)

    (Original post by KloppOClock)
    The fundamental problem is that you are applying some function f: \mathbb{R} \to \mathbb{R} to both sides of the original equation, to create a new equation. You require that the new equation and the original equation are equivalent i.e. that they have the same set of solutions.

    To express this logically, you require the following:

    a=b \Leftrightarrow f(a)=f(b)

    Now the forward implication a=b \Rightarrow f(a)=f(b) is fine , since that is true as part of the definition of a function (they have to be single-valued), but the reverse implication f(a)=f(b) \Rightarrow a=b says that f must be one-one, and some functions aren't e.g:

    f(x)=x^2, f(1)=1^2=1, f(-1)=(-1)^2=1
    f(x)=\sin(x), f(a)=\sin(a), f(a+2\pi)=\sin(a+2\pi)=\sin(a)

    and so on. So you must only transform your equation using functions that are both one-one and onto (since the function must be defined for all possible values of both sides of the original), and that means that f must be invertible i.e. f^{-1} must exist.

    1. The first problem is that if you don't use an invertible function you can create extraneous solutions. Logically, you have something like this, for f(x)=x^2:

    (x=a \Rightarrow f(x)=a^2) \land (x=-a \Rightarrow f(x)=a^2)


    f(x)=a^2 \Rightarrow (x=a) \lor (x=-a)

    so if we start with the trivial equation x=1 and apply f to both sides, we must write:

    x=1 \Rightarrow x^2 =1 \Rightarrow (x=1) \lor (x=-1)

    which has produced the extraneous solution x=-1.

    We can fix this up, however, by carrying across information from the original equation into the transformed equation. For example, note that f(x)=\sqrt{x} has range [0,\infty). We can then do, say, the following:

    1. Start with some equation: \sqrt{x+5}=x+3.
    2. Here we need x+3 = \sqrt{x+5} \ge 0 \Rightarrow x \ge -3
    3. Add the currently unstated range restriction: (\sqrt{x+5}=x+3) \land (x \ge -3)
    4. Solve the equation by squaring:

    (\sqrt{x+5}=x+3) \land (x \ge -3)  \\ \Leftrightarrow  (x+5=x^2+6x+9) \land (x \ge -3)  \\ \Leftrightarrow   (x^2+5x+4=0) \land (x \ge -3)  \\ \Leftrightarrow  (x+4)(x+1)=0 \land (x \ge -3) \\ \Leftrightarrow  (x=-4) \lor (x=-1) \land (x \ge -3)  \\ \Leftrightarrow  x=-1

    where, in the final step, the logical and of x=-4 and x \ge -3 removes the extraneous solution. Note that I can use \Leftrightarrow now that I've added in the range restriction, so we can see that the final equation and original equation are completely equivalent. (Note also that although there were several transformations applied to various equtations, only the first is due to a non-invertible function, so we don't have to worry about the rest.)

    A common problem is when we clear the denominator of a fractional expression, since f_x(a) = ax is only invertible if x \ne 0 (since e.g. f_0(1)=1 \times 0 =0, f_0(2) = 2 \times 0 = 0).

    So e.g. we may have to do, say, the following:

    \displaystyle \frac{\cos x}{\sin x+1} = 0 \\ \Leftrightarrow \cos x =0 \land (\sin x +1 \ne 0) \\ \Leftrightarrow x \in \{ \frac{\pi}{2}, \frac{3\pi}{2},  \frac{5\pi}{2}, \cdots \} \land x \notin \{\frac{3\pi}{2},  \frac{7\pi}{2}, \cdots \} \\ \Leftrightarrow x \in \{   \frac{\pi}{2},   \frac{5\pi}{2}, \cdots \}

    so again the original and final equations have the same set of solutions, once I've added the required restriction on the denominator as a logical condition, which ensures invertibility of all operations.

    2. We can also have the problem of losing solutions when applying a function to an equation e.g.

    x^2+x=0 \Rightarrow x(x+1)=0 \Rightarrow x+1=0 \Rightarrow x=-1

    We lost the solution x=0 by dividing away the x. What's the problem here?

    Well, the function that we applied to both sides here is f(a) = \frac{a}{x}, and that domain of that function does not include x=0. So by dividing by x, we are implicitly saying:

    x(x+1)=0 \land x \ne 0 \\ \Leftrightarrow f(x(x+1)) = f(0)  \\ \Leftrightarrow x+1=0 \land x \ne 0 \\ \Leftrightarrow x=-1 \land x \ne 0

    i.e. to allow us to divide away x, we immediately have to remove one of the values that happens to be a solution to the original equation.
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