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# Solving Quadratic Equations with Square Roots

## Solving Quadratic Equations with Square Roots 1.9 - Solution

a

We want to find the domain of the given quadratic function. $\begin{gathered} f(x)=x^2-9 \end{gathered}$ We can find its domain by graphing it. To do so, let's first make a table of values.

$x$ $x^2-9$ $y=x^2-9$
${\color{#0000FF}{\text{-}3}}$ $({\color{#0000FF}{\text{-}3}})^2-9$ $0$
${\color{#0000FF}{\text{-}2}}$ $({\color{#0000FF}{\text{-}2}})^2-9$ $\text{-}5$
${\color{#0000FF}{0}}$ ${\color{#0000FF}{0}}^2-9$ $\text{-}9$
${\color{#0000FF}{2}}$ ${\color{#0000FF}{2}}^2-9$ $\text{-}5$
${\color{#0000FF}{3}}$ ${\color{#0000FF}{3}}^2-9$ $0$

Next, we will plot the obtained points on the coordinate plane and connect them with a parabola.

We see above that the $x\text{-}$variable can take any value. With this in mind, we can state the domain. $\begin{gathered} \textbf{Domain:}\ \text{all real numbers} \end{gathered}$

b

To find the range of the function, we will again interpret its graph.

As we can see, the $y\text{-}$variable only takes values greater than or equal to $\text{-}9.$ $\begin{gathered} \textbf{Range:}\ y\geq \text{-} 9 \end{gathered}$

c

Looking at the graph, we can conclude that for $x\text{-}$values between the $x\text{-}$intercepts of the parabola, the value of $f(x)$ is negative.

Let's find the $x\text{-}$intercepts of the graph! To do so, we substitute $0$ for $f(x)$ in the function and solve for $x.$
$f(x)=x^2-9$
${\color{#0000FF}{0}}=x^2-9$
$9=x^2$
$\pm3=x$
$x=\pm3$
As a result, $f(x)$ is negative for $x\text{-}$values between $\text{-}3$ and $3.$ $\begin{gathered} \text{-}3 < x < 3 \end{gathered}$