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anonymous

  • one year ago

Given the equation −4Square root of x minus 3 = 12, solve for x and identify if it is an extraneous solution. x = 0, solution is not extraneous x = 0, solution is extraneous x = 12, solution is not extraneous x = 12, solution is extraneous

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  1. anonymous
    • one year ago
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    x=12, solution is extraneous

  2. anonymous
    • one year ago
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    kthx ;)

  3. anonymous
    • one year ago
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    If you wanted an explanation, go about this like a normal algebraic explanation (aka try to find the value of x and don't worry about the extraneous part yet).\[-4\sqrt{x-3} = 12\]\[\sqrt{x-3}=-3~~~\text{<---Take note of this, we'll come back here later.}\]\[x - 3=9\]\[x=12\]Ok, great. We got a solution. Now, the term extraneous solution means that the solution that we found doesn't actually work. Let's test the solution by plugging it back in.\[-4\sqrt{x-3} = 12\]\[-4\sqrt{12-3}=12\]\[-4\sqrt{9}=12\]\[-12\ne12\]Since -12 doesn't equal 12, the solution doesn't work. Now, if you go back to that second step, You might've noticed that \(\sqrt{x-3}\) can NEVER BE NEGATIVE, yet it is set equal to a negative number. (when a \(\sqrt{}\) is given to you, it is assumed positive. So, \(\sqrt{x^2} = |x|\), not x because it can never be negative. However, if you take the squareroot to solve an equation, like \(x^2=9\), then you get \(x=\pm3\) because you're trying to find all possible solutions.) This is something that would immediately tip you off that the solution you WILL get is extraneous, because this is impossible to satisfy.

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