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The rectangular board in LaShay's class is $\sqrt{x+5}$ wide feet by $8$ feet long.

After an exhausting school day, LaShay went to an amusement park to have fun with her friends. The first thing they did was ride on a Ferris wheel.

After an exciting day at the amusement park, even if the math did not help LaShay save her phone from smashing against the ground, her interest in radical equations increased. Right now, LaShay and Ali are trying to solve the following radical equation.

After some work, they managed to solve the equation. However, they came up with different solutions.

After seeing LaShay's hard work, her teacher gives her extra credit homework so that she can have more practice with radical equations.

Solution

To solve a radical equation graphically, both sides of the equation need to be represented by functions.

$$\begin{array}{c}\sqrt{2x-6}=1+\sqrt{x-2}\\ \Downarrow \\ y=\sqrt{2x-6}\text{ and }y=1+\sqrt{x-2}\end{array}$$

Now that there are two radical functions, they can be graphed by making a table of values. Since the radicands cannot be negative, the domain of each function will be found.

$$\begin{array}{cc}\text{Domain of}& \text{Domain of}\\ \mathbf{y}=\sqrt{2\mathbf{x}-6}& \mathbf{y}=1+\sqrt{\mathbf{x}-2}\\ 2x-6\ge 0& x-2\ge 0\\ \Updownarrow & \Updownarrow \\ x\ge 3& x\ge 2\end{array}$$

Therefore, the table of values for the first function will be made for values of $x$ greater than or equal to $3.$ For the second function, the $x\text{-}$values must be greater than or equal to $2.$

	$y=\sqrt{2x-6}$		$y=1+\sqrt{x-2}$
$x$	$\sqrt{2x-6}$	$y$	$1+\sqrt{x-2}$	$y$
$2$	-	-	$1+\sqrt{2-2}$	$1$
$3$	$\sqrt{2(3)-6}$	$0$	$1+\sqrt{3-2}$	$2$
$5$	$\sqrt{2(5)-6}$	$2$	$1+\sqrt{5-2}$	$\approx 2.73$
$6$	$\sqrt{2(6)-6}$	$\approx 2.45$	$1+\sqrt{6-2}$	$3$
$11$	$\sqrt{2(11)-6}$	$4$	$1+\sqrt{11-2}$	$4$

Now the points found in both functions can be plotted in the same coordinate plane. Then they can be connected with smooth curves.

Finally, the $x\text{-}$coordinate of the point of intersection of the curves gives the solution of the original equation.

Because the point of intersection is at $(11,4),$ the solution to the equation is $x=11.$

On Monday, LaShay's math teacher asks her how her extra credit homework went. To show what she has learned, LaShay gives him the following radical expression, which represents the number of hours that she studied over the weekend.

Solution

Since the inequality phrase is less than, the inequality symbol is $<.$ With this information, the inequality that represents the situation can be written.

$$\begin{array}{c}\sqrt{2x-4}+4<8\end{array}$$

Note that the index of the root is $2,$ which is an even number. Therefore, the radicand must be non-negative.

$2x-4\ge 0$

AddIneq

$\text{LHS}+4\ge \text{RHS}+4$

$2x\ge 4$

DivIneq

$\text{LHS}/2\ge \text{RHS}/2$

$x\ge 2$

It was found that $x$ is greater than or equal to $2.$ Next isolate the radical expression in the original inequality and solve for $x.$

$\sqrt{2x-4}+4<8$

▼

Solve for $x$

SubIneq

$\text{LHS}-4<\text{RHS}-4$

$\sqrt{2x-4}<4$

RaiseIneq

${\text{LHS}}^2<{\text{RHS}}^2$

${\left(\sqrt{2x-4}\right)}^2<16$

PowSqrt

${\left(\sqrt{a}\right)}^2=a$

$2x-4<16$

AddIneq

$\text{LHS}+4<\text{RHS}+4$

$2x<20$

DivIneq

$\text{LHS}/2<\text{RHS}/2$

$x<10$

The values of $x$ are also less than $10.$ By combining both statements, the solution set can be written.

$$\begin{array}{c}x\ge 2\text{and}x<10\\ \Updownarrow \\ 2\le x<10\end{array}$$

Finally, some test values will be checked in the original inequality to confirm the solution. Use three test values, one value less than $2,$ one value greater than or equal to $2$ and less than $10,$ and one value greater than or equal to $10.$ In this case the test values $0,$ $4,$ and $20$ will be verified.

$\sqrt{2x-4}+4<8$
$x$	Substitute	Simplify
$0$	$\sqrt{2(0)-4}+4<8$	$\sqrt{\text{-}4}\nless 4\times$
$4$	$\sqrt{2(4)-4}+4<8$	$6<8✓$
$20$	$\sqrt{2(20)-4}+4<8$	$10\nless 8\times$

The values that do not belong to the solution set, $0$ and $20,$ do not satisfy the inequality. Conversely, the value that does belong to the solution set, $4,$ satisfies the inequality. Therefore, the solution checks out. The solution set can also be graphed on a number line.

The challenge given at the beginning can be solved using the methods covered throughout the lesson. Recall that LaShay and Ali will participate in the upcoming Mountain Bike Race.