a \(y + \sqrt {y - 4} = 4\)

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In this case let’s notice that if we just square both sides we’re going to have problems.

\[\begin{align*}{\left( {y + \sqrt {y - 4} } \right)^2} & = {\left( 4 \right)^2}\\ {y^2} + 2y\sqrt {y - 4} + y - 4 & = 16\end{align*}\]

Before discussing the problem we’ve got here let’s make sure you can do the squaring that we did above since it will show up on occasion. All that we did here was use the formula

\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]

with \(a = y\) and \(b = \sqrt {y - 4} \). You will need to be able to do these because while this may not have worked here we will need to this kind of work in the next set of problems.

Now, just what is the problem with this? Well recall that the point behind squaring both sides in the first problem was to eliminate the square root. We haven’t done that. There is still a square root in the problem and we’ve made the remainder of the problem messier as well.

So, what we’re going to need to do here is make sure that we’ve got a square root all by itself on one side of the equation before squaring. Once that is done we can square both sides and the square root really will disappear.

Here is the correct way to do this problem.

\[\begin{align*}\sqrt {y - 4} & = 4 - y\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{\mbox{now square both sides}}\\ {\left( {\sqrt {y - 4} } \right)^2} & = {\left( {4 - y} \right)^2}\\ y - 4 & = 16 - 8y + {y^2}\\ 0 & = {y^2} - 9y + 20\\ 0 & = \left( {y - 5} \right)\left( {y - 4} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}y = 4,\,\,\,y = 5\end{align*}\]

As with the first example we will need to make sure and check both of these solutions. Again, make sure that you check in the original equation. Once we’ve square both sides we’ve changed the problem and so checking there won’t do us any good. In fact, checking there could well lead us into trouble.

First \(y = 4\).

\[\begin{align*}4 + \sqrt {4 - 4} & \mathop = \limits^? 4\\ 4 & = 4\hspace{0.25in}{\mbox{OK}}\end{align*}\]

So, that is a solution. Now \(y = 5\).

\[\begin{align*}5 + \sqrt {5 - 4} & \mathop = \limits^? 4\\ 5 + \sqrt 1 & \mathop = \limits^? 4\\ 6 & \ne 4\hspace{0.25in}{\mbox{NOT }}{\mbox{OK}}\end{align*}\]

So, as with the first example we worked there is in fact a single solution to the original equation, \(y = 4\).

b \(1 = t + \sqrt {2t - 3} \)

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Okay, so we will again need to get the square root on one side by itself before squaring both sides.

\[\begin{align*}1 - t & = \sqrt {2t - 3} \\ {\left( {1 - t} \right)^2} & = {\left( {\sqrt {2t - 3} } \right)^2}\\ 1 - 2t + {t^2} & = 2t - 3\\ {t^2} - 4t + 4 & = 0\\ {\left( {t - 2} \right)^2} & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}t = 2\end{align*}\]

So, we have a double root this time. Let’s check it to see if it really is a solution to the original equation.

\[\begin{align*}1 & \mathop = \limits^? 2 + \sqrt {2\left( 2 \right) - 3} \\ 1 & \mathop = \limits^? 2 + \sqrt 1 \\ 1 & \ne 3\end{align*}\]

So, \(t = 2\) isn’t a solution to the original equation. Since this was the only possible solution, this means that there are **no solutions** to the original equation. This doesn’t happen too often, but it does happen so don’t be surprised by it when it does.

c \(\sqrt {5z + 6} - 2 = z\)

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This one will work the same as the previous two.

\[\begin{align*}\sqrt {5z + 6} & = z + 2\\ {\left( {\sqrt {5z + 6} } \right)^2} & = {\left( {z + 2} \right)^2}\\ 5z + 6 & = {z^2} + 4z + 4\\ 0 & = {z^2} - z - 2\\ & 0 = \left( {z - 2} \right)\left( {z + 1} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,z = - 1,\,\,\,\,z = 2\end{align*}\]

Let’s check these possible solutions start with \(z = - 1\).

\[\begin{align*}\sqrt {5\left( { - 1} \right) + 6} - 2 & \mathop = \limits^? \, - 1\\ \sqrt 1 - 2 & \mathop = \limits^? \; - 1\\ - 1 & = - 1\hspace{0.25in}{\mbox{OK}}\end{align*}\]

So, that’s was a solution. Now let’s check \(z = 2\).

\[\begin{align*}\sqrt {5\left( 2 \right) + 6} - 2 & \mathop = \limits^? 2\\ \sqrt {16} - 2 & \mathop = \limits^? 2\\ 4 - 2 & = 2\hspace{0.25in}{\mbox{OK}}\end{align*}\]

This was also a solution.

So, in this case we’ve now seen an example where both possible solutions are in fact solutions to the original equation as well.