To solve for y, we first want to isolate \sqrt{y}, we can then square both sides of the equation.

We can see that if we substitute y=16 into the original equation we obtain \sqrt{16}+5=9 which is a true statement, and as \sqrt{16} is positive we have a valid solution.

To solve for x, we first want to isolate 4x-9 by raising each side of the equation to a power of 3. We are then left with a linear equation we can solve using more inverse operations

We can see that if we substitute x=2 into the original equation we obtain \sqrt[3]{-1}=-1 which is a true statement. When dealing with odd valued indexes such as \sqrt[3]{x}, we do not need to worry about the radicand being negative.

Raising both sides of an equation to an odd power is reversible. If a=b then a^3=b^3 and if a^3=b^3 then a=b.

To solve for x, we first want to isolate 2\sqrt{x}. We can then square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

Using the zero product rule we get two solutions x=\dfrac{1}{9}, x=1. We now want to test both solutions:

We can see that x=1 satisfies the original equation and is a valid solution.

Now, let's test x=\dfrac{1}{9}

We can see that x=\dfrac{1}{9} leads to a false statement and is therefore an extraneous solution.

To solve for x, we first want to square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

Using the zero product rule we get two solutions x=-8, x=-1. We now want to test both solutions:

We can see that x=-8 leads to a false statement and is therefore an extraneous solution.

Now, let's test x=-1.

We can see that x=-1 satisfies the original equation and is a valid solution.

It is important to note that the fact that our solutions were negative, x=-8, x=-1, does not necessarily make them extraneous. We can see after substitution that only x=- 8 is an extraneous solution, and x=-1 is valid.

1. Consider the inequality \sqrt{3x+12} \lt 6 and isolate x using inverse operations.

2. Consider the additional condition that the radicand must be positive by solving the inequality 3x+12 \geq 0 for x.

3. Find the solution of the intersection of the two intervals found and represent the solution graphically.

4. Check the solution set includes all viable solutions.

Step 1. Consider the inequality \sqrt{3x+12} \lt 6 and isolate x:

Step 2. Consider the inequality 3x+12 \geq 0 and isolate x:

Step 3. The solution set will be all the numbers that make both inequalities true, that is, both x \geq -4 and x \lt 8. This can be written as a single inequality as -4 \leq x \lt 8 and represented graphically on a number line as follows:

Step 4. Notice the two points x=-4 and x=8 divide the number line into three sections. Numbers less than -4 will make the radicand negative, so cannot be part of the solution set. We can double check the correct interval is highlighted by checking a point inside the other two sections to determine if it is part of the solution set. For example, we can test x=0 and x=10.

We can see that x=0 leads to a true statement and is therefore a valid solution along with other points in the interval -4 \leq x\lt 8.

We can see that x=10 leads to a false statement and is therefore not a valid solution along with other points in the interval x\geq 8.

When creating the number line note the signs of the inequality and use an unfilled circle when an endpoint is not included and a filled circle when an endpoint is included.

As squaring both sides of an equation or inequality may introduce extraneous solutions it is important to check we have the correct solution set.

To solve for x, we first want to isolate the radical term and then cube both sides of the inequality to remove the radical. We can then solve the resulting linear inequality.

This can be represented graphically on a number line as follows:

For roots with an odd index, we don’t have additional conditions for what is underneath the radical sign, since we could have positive or negative numbers as a radicand.

To solve using technology graph f\left(x\right)=\sqrt{3x-6}+1 and g\left(x\right)=2x-6 and determine when the function f\left(x\right) is equal or above the function g\left(x\right).

Enter the left side of the inequality as f\left(x\right) and the right side as g\left(x\right). Then use the intersection tool to find when the two functions are equal.

We can see f\left(x\right) has a domain of x\geq 2 and lies above g\left(x\right) on the interval 2 \leq x \leq 5.

This can be represented graphically on a number line as follows:

Technology can assist to solve multi-step inequalities and to confirm the solution set when solving algebraically.

We could also have solved the above inequality algebraically as follows:

Consider the inequality \sqrt{3x-6}+1 \geq 2x-6 and isolate x:

We can visualize the solution set by considering the graph of the corresponding quadratic equation:

Then, consider the inequality representing the restriction on the radicand 3x-6 \geq 0 and isolate x:

For inequalities when there is a variable not under a square root and we square both sides, we have to be particularly careful to check solutions. Check the interval by testing values in the original inequality.

Notice the three points x=2, x=\dfrac{11}{4}, and x=5 divide the number line into four sections. Numbers less than 2 will make the radicand negative, so cannot be part of the solution set. We can find the solution set by checking a point inside the other three sections to determine if it is part of the solution set. For example, we can test x=2.5, x=4 and x=6.

Checking these points in the inequality \sqrt{3x-6}+1 \geq 2x-6 creates a true statement for {x=2.5}, x=4 and a false statement for x=6. Thus, the solution set is: 2 \leq x \leq 5.