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Solving Quadratic Equations with Square Roots

Solving Quadratic Equations with Square Roots 1.9 - Solution

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a

We want to find the domain of the given quadratic function. f(x)=x29\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.

xx x29x^2-9 y=x29y=x^2-9
-3{\color{#0000FF}{\text{-}3}} (-3)29({\color{#0000FF}{\text{-}3}})^2-9 00
-2{\color{#0000FF}{\text{-}2}} (-2)29({\color{#0000FF}{\text{-}2}})^2-9 -5\text{-}5
0{\color{#0000FF}{0}} 029{\color{#0000FF}{0}}^2-9 -9\text{-}9
2{\color{#0000FF}{2}} 229{\color{#0000FF}{2}}^2-9 -5\text{-}5
3{\color{#0000FF}{3}} 329{\color{#0000FF}{3}}^2-9 00

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

We see above that the x-x\text{-}variable can take any value. With this in mind, we can state the domain. Domain: all real numbers\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-y\text{-}variable only takes values greater than or equal to -9.\text{-}9. Range: y-9\begin{gathered} \textbf{Range:}\ y\geq \text{-} 9 \end{gathered}

c

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

Let's find the x-x\text{-}intercepts of the graph! To do so, we substitute 00 for f(x)f(x) in the function and solve for x.x.
f(x)=x29f(x)=x^2-9
0=x29{\color{#0000FF}{0}}=x^2-9
9=x29=x^2
±3=x\pm3=x
x=±3x=\pm3
As a result, f(x)f(x) is negative for x-x\text{-}values between -3\text{-}3 and 3.3. -3<x<3\begin{gathered} \text{-}3 < x < 3 \end{gathered}