Algebra - Solutions and Solution Sets

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Section 2.1 : Solutions and Solution Sets

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4. Is \(w = - 2\) a solution to \(\frac{{{w^2} + 8w + 12}}{{w + 2}} = 0\)?

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There really isn’t all that much to do for these kinds of problems. All we need to do is plug the given number into both sides of the equation and check to see if the right and left side are the same value.

Note that for this problem we don’t even really need to plug the value into the equation. We can see by a quick inspection that if we were to plug \(w = - 2\) into this equation we would have division by zero and we know that is not allowed.

Therefore, \(w = - 2\) is not a solution to this equation.

Be very careful with this kind of problem. If we had plugged \(w = - 2\) into the equation we’d have gotten zero in the numerator as well and we might be tempted to say that it is a solution to the equation. We’d be wrong however. Regardless of the value of the numerator, we would still have division by zero and that is just not allowed and so \(w = - 2\) will not be a solution to this equation.