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Section 2.5 : Quadratic Equations - Part I

13. Use the Square Root Property to solve the equation.

\[9{u^2} - 16 = 0\]

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There really isn’t too much to this problem. Just recall that we need to get the variable on one side of the equation by itself with a coefficient of one. For this problem that gives,

\[\begin{align*}9{u^2} & = 16\\ {u^2} & = \frac{{16}}{9}\end{align*}\] Show Step 2

Now all we need to do is use the Square Root Property to get,

\[u = \pm \sqrt {\frac{{16}}{9}} = \pm \frac{{\sqrt {16} }}{{\sqrt 9 }} = \pm \frac{4}{3}\]

So we have the following two solutions : \(\require{bbox} \bbox[2pt,border:1px solid black]{{u = - \frac{4}{3}\,\,\,{\mbox{and }}u = \frac{4}{3}}}\).