Solving Radical Inequalities

Solve $2+\sqrt{4x-4}\le 6$

Solution: To solve this radical inequality, let's follow all the steps.

Step 1: First we check the index of the given radical inequality. As no index value is given, the index value is 2.

Step 2: As the index is even, the radicand of the square root will be greater than or equal to zero.

$$\begin{gathered} 4x-4\ge 0 \\ \Rightarrow 4x\ge 4 \\ \Rightarrow x\ge 1\left(1\right) \end{gathered}$$

Step 3: Now we will solve the radical inequality algebraically and also remove the radical symbol to simplify it. First, we isolate the radical.

$2+\sqrt{4x-4}\le 6\Rightarrow \sqrt{4x-4}\le 4$

Now, we eliminate the radical symbol by taking index as an exponent on both sides of the inequality.

$$\begin{gathered} {\left(\sqrt{4x-4}\right)}^2\le 4^2 \\ \Rightarrow 4x-4\le 16 \\ \Rightarrow 4x\le 20 \\ \Rightarrow x\le 5\left(2\right) \end{gathered}$$

Here, we got two inequalities for the value of x from equation $\left(1\right)$ and $\left(2\right)$. So we combine them both and write it as a compound inequality. So our final answer is:

$1\le x\le 5$

Here note that 1 is the lower interval value and 5 is the upper interval value.

Step 4: Finally, we will test some values of x to check our solution and confirm it. Let's consider $x=0,8$ outside our range of x and $x=3$ inside our range of x.

We can see that the value of x satisfies the radical inequality. So the solution $\mathbf{1}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{5}$ checks and satisfies the given radical inequality.

Solve $\sqrt{x-5}>3$

Solution:

Step 1: We consider $y=\sqrt{x-5}$ and $y=3$.

Step 2: Now we plot the graph for both the functions from step 1.

Radical inequality graph, Mouli Javia - StudySmarter Originals

Here, the graph with the red line is of the function $y=\sqrt{x-5}$and the graph with the green line is of the function $y=3$. We have plotted the graph such that we can clearly identify the x-axis values between 0 and 30. Similarly, we can clearly take note of the values on the y-axis.

Step 3: Now, we identify the values of x for which the first graph $y=\sqrt{x-5}$ lies above the second graph $y=3$, as we have a greater than inequality sign in the original inequality equation. We can also see that value of x intersects both the graphs at $x=14$. This implies that the first graph lies above the second graph for $x>14.$

$\therefore$The solution to this radical inequality is $\mathit{x}\mathbf{>}\mathbf{14}$.

Solve $\sqrt[3]{x+2}\ge 1,$ algebraically and graphically.

Solution: First we solve using the algebra method.

Step 1: First we check the index of the given radical inequality. Here it is 3.

Step 2: As the index is odd, we consider: $x+2\ge 1\Rightarrow x\ge -1.$

Step 3: Now we solve the original inequality. As the given inequality does not have other operations, we skip the step of isolation.

$$\begin{gathered} \sqrt[3]{x+2}\ge 1 \\ \Rightarrow {\left(\sqrt[3]{x+2}\right)}^3\ge 1^3 \\ \Rightarrow x+2\ge 1 \\ \therefore x\ge -1 \end{gathered}$$

So the values of x from step 2 and step 3 are $x\ge -1$.

Step 4: Now we check our solution to confirm it.

Hence, it can be seen that the given radical inequality is satisfied for the values$\mathit{x}\mathbf{\ge }\mathbf{-}\mathbf{1}$.

Now we will solve the same radical inequality using the graphical method.

Step 1: We consider $y=\sqrt[3]{x+2}$ and $y=1.$

Step 2: we plot the graphs for two functions taken into account in step 1.

graphical form of radical inequality, Mouli Javia - StudySmarter Originals

In the above graph, the red line represents the function $y=\sqrt[3]{x+2}$ and the blue line represents the function$y=1$.

Step 3: We identify the values of x for which the first graph $y=\sqrt[3]{x+2}$ lies above the graph $y=1$. And the value of x intersects both the graphs at $x=-1$. So the first graph lies above the second graph for the values $x\ge -1$.

Hence, the solution for the given radical inequality is $\mathit{x}\mathbf{\ge }\mathbf{-}\mathbf{1}$.