How to Solve Radical Inequalities?

We can solve radical inequalities using algebra. Some radical inequalities also have variables outside the radical, and we can use algebra to calculate them as well. The following steps can be used to solve radical inequalities:

Step 1: Check the index of the radical.

• If the index is even, the radicand cannot be negative, so it must be greater than or equal to zero. This is called domain restriction.

Step 2: If the index is even, treat the radicand as greater than or equal to zero. Solve for the variable x in radicands.

• Therefore, we solve for the variable \(x\) for this radicand when it is greater than or equal to zero. That is, we consider the radicand as \(x\ge 0\) from the radical inequality \(\sqrt[n]{x}<d\) and calculate the variable \(x\). If the index is odd, however, the radicand may be negative, so there is no domain restriction: raise both sides to the \(n\)th power to get \(x<d^n\).

Step 3: Solve the original inequality expression algebraically and also remove the radical symbol from the expression.

• We eliminate the radical by taking the index and using it as the exponent in terms of both sides of the inequality. (i.e., \(\sqrt[n]{x}<d\:\rightarrow \:\left(\sqrt[n]{x}\right)^n<d^n)\). Note here that when using the index as an exponent on the radical expression, it nullifies the radical symbol, thus removing it.

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Step 4: Test the values to check the solution.

• To test the values of \(x\), we consider a random value that satisfies the inequality. And we will also consider values outside the equality so that we can confirm the correctness of our solution.

Solving Radical Inequalities , Example 1:

solve \(3+\sqrt{4x-4}\le 7\).

Solution:

To solve this radical inequality, first, we check the index of the given radical inequality. Since the index value is not given, the index value is \(2\). Since the index is even, the radicand of the square root will be greater than or equal to zero.

\(4x-4\ge 0\)

\(4x\ge 4\)

\(x\ge 1\)…………. \((1)\)

We now solve the radical inequality algebraically and also remove the radical symbol to simplify it. First, we isolate the radical.

\(3+\sqrt{4x-4}\le 7\rightarrow \sqrt{4x-4}\le 4\)

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

\(\left(\sqrt{4x-4}\right)^2\le 4^2\)

\(4x-4\le 16\)

\(4x\le 20\)

\(x\le 5\)…………. \((2)\)

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

\(1\le x\le 5\)

Exercises for Solving Radical Inequalities

Solve.

1. \(\color{blue}{\sqrt[3]{x+3}\ge \:2}\)
2. \(\color{blue}{-2\sqrt{x+1}\le -6}\)
3. \(\color{blue}{4\sqrt[3]{x+1}\ge 12}\)

1. \(\color{blue}{x\ge 5}\)
2. \(\color{blue}{\:x\ge 8}\)
3. \(\color{blue}{x\ge 26}\)

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Special Cases in Radical Inequalities

A square root is never negative. That one fact settles some problems before you do any algebra.

• Radical less than a negative number: √(x − 1) < −2 has no solution, because √(x − 1) is always 0 or more.
• Radical greater than a negative number: √(x − 1) ≥ −2 is true for every x in the domain, so the answer is x ≥ 1.
• Odd roots: a cube root can be negative, so there is no domain step. For ∛(x + 1) < −2, cube both sides: x + 1 < −8, so x < −9.

Radicals on both sides

Solve √(2x − 1) > √(x + 3). Both radicands must be 0 or more: 2x − 1 ≥ 0 gives x ≥ ½, and x + 3 ≥ 0 gives x ≥ −3. Together, x ≥ ½. Now square both sides: 2x − 1 > x + 3, so x > 4. Since x > 4 is inside the domain, the answer is x > 4.

Radical Inequality with a Variable Outside the Radical

Solve √(x + 2) ≤ x − 10.

1. Domain: x + 2 ≥ 0, so x ≥ −2.
2. The right side must be 0 or more, since a square root cannot be less than a negative: x − 10 ≥ 0, so x ≥ 10.
3. Square both sides: x + 2 ≤ x² − 20x + 100.
4. Rearrange: 0 ≤ x² − 21x + 98, which factors as (x − 7)(x − 14) ≥ 0. So x ≤ 7 or x ≥ 14.
5. Combine with x ≥ 10: the answer is x ≥ 14.

Check: x = 14 gives √16 = 4 ≤ 4, true. x = 12 gives √14 ≈ 3.74 ≤ 2, false. Testing one value on each side of every boundary is the quickest way to catch a mistake.

Frequently Asked Questions

How do you solve radical inequalities?

To solve a radical inequality, first find the domain, then isolate the radical and raise both sides to the power of the index. Solve the new inequality and keep only the values that also satisfy the domain. For example, √(x + 4) < 3 needs x ≥ −4. Squaring gives x + 4 < 9, so x < 5. The answer is −4 ≤ x < 5.

How do you set up an inequality showing that the radicand cannot be negative?

Take the expression under the even root and set it greater than or equal to zero. For √(3x − 6), the condition is 3x − 6 ≥ 0, which gives x ≥ 2. This domain inequality comes first, because the square root of a negative number is not a real number. Your final answer must stay inside this domain.

What is the difference between radical equations and radical inequalities?

Both are solved by isolating the radical and raising both sides to the index power. A radical equation usually gives one or two values to check, while a radical inequality gives a whole range of values. For example, √x = 4 gives x = 16, but √x ≤ 4 gives 0 ≤ x ≤ 16 once you include the domain x ≥ 0.

How do you solve inequalities with square roots?

When a square root is greater than a positive number, square both sides and combine the result with the domain. For √(2x + 1) > 3, the domain is x ≥ −1/2. Squaring gives 2x + 1 > 9, so 2x > 8 and x > 4. Every x > 4 is already in the domain, so the solution is simply x > 4.

Is there a solving radical inequalities worksheet?

Yes, a solving radical inequalities worksheet that mixes square root and cube root problems is great practice. Work each problem in the same order: domain, isolate, raise to the power, solve, then check a test value. For √(x − 1) ≤ 2, the answer is 1 ≤ x ≤ 5. Try x = 3: √2 ≈ 1.41 is less than 2, so it checks out.