College Algebra, Chapter 1, 1.5, Section 1.5, Problem 32

Find all real solutions of the equation $\displaystyle 2x + \sqrt{x + 1} = 8$


$
\begin{equation}
\begin{aligned}

2x + \sqrt{x + 1} =& 8
&& \text{Given}
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\sqrt{x + 1} =& 8 - 2x
&& \text{Subtract } 2x
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(\sqrt{x + 1})^2 =& (8 - 2x)^2
&& \text{Square both sides}
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x + 1 =& 64 - 32x + 4x^2
&& \text{Combine like terms}
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4x^2 - 33x + 63 =& 0
&& \text{Subtract } 63
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4x^2 - 33x =& -63
&& \text{Divide both sides by } 4
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x^2 - \frac{-33x}{4} =& \frac{-63}{4}
&& \text{Complete the square: add } \left( \frac{\displaystyle \frac{-33}{4}}{2} \right)^2 = \frac{1089}{64}
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x^2 - \frac{33x}{4} + \frac{1089}{64} =& \frac{-63}{4} + \frac{1089}{64}
&& \text{Perfect square}
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\left(x - \frac{33}{8} \right)^2 =& \frac{81}{64}
&& \text{Take the square root}
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x - \frac{33}{8} =& \pm \sqrt{\frac{81}{64}}
&& \text{Add } \frac{33}{8} \text{ and simplify}
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x =& \frac{33}{8} \pm \frac{9}{8}
&& \text{Evaluate}
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x =& \frac{21}{4} \text{ and } x = 3
&& \text{Solve for } x
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x =& 3
&& \text{The only solution that satisfy the equation } 2x \sqrt{x + 1} = 8

\end{aligned}
\end{equation}
$