A graded collection of 20 worked exercises on irrational inequalities, designed to develop a thorough understanding of how to solve inequalities involving radicals, with careful attention to domain conditions, the sign of each side, and the identification of extraneous solutions.
In each exercise we do not simply square both sides and move on — we justify why that step is valid. This is the central issue with irrational inequalities: squaring is not always an equivalent transformation unless the appropriate conditions are checked first.
Exercise 1 — level ★☆☆☆☆
Solve:
\[ \sqrt{x-2}\gt 3. \]
Answer
\[ S=(11,+\infty). \]
Solution
The radical is defined when:
\[ x-2\ge 0, \]
that is:
\[ x\ge 2. \]
Since the right-hand side is positive, we may square both sides:
\[ x-2\gt 9. \]
Hence:
\[ x\gt 11. \]
Intersecting with the domain yields:
\[ S=(11,+\infty). \]
Exercise 2 — level ★☆☆☆☆
Solve:
\[ \sqrt{2x+1}\le 5. \]
Answer
\[ S=\left[-\frac{1}{2},12\right]. \]
Solution
The domain condition is:
\[ 2x+1\ge 0, \]
that is:
\[ x\ge -\frac{1}{2}. \]
The right-hand side is positive, so we may square both sides:
\[ 2x+1\le 25. \]
Hence:
\[ 2x\le 24, \]
so:
\[ x\le 12. \]
Intersecting the two conditions:
\[ S=\left[-\frac{1}{2},12\right]. \]
Exercise 3 — level ★☆☆☆☆
Solve:
\[ \sqrt{x+4}\lt 2. \]
Answer
\[ S=[-4,0). \]
Solution
The radical is defined when:
\[ x+4\ge 0, \]
that is:
\[ x\ge -4. \]
The right-hand side is positive, so we may square both sides:
\[ x+4\lt 4. \]
Hence:
\[ x\lt 0. \]
Intersecting with the domain:
\[ S=[-4,0). \]
Exercise 4 — level ★★☆☆☆
Solve:
\[ \sqrt{x+1}\ge x-1. \]
Answer
\[ S=[-1,3]. \]
Solution
The radical is defined when:
\[ x+1\ge 0, \]
that is:
\[ x\ge -1. \]
The left-hand side is always nonnegative. We therefore examine the sign of the right-hand side \(x-1\).
Case 1: \(x-1\le 0\)
If:
\[ x-1\le 0, \]
then \(x\le 1\). In this case the right-hand side is nonpositive while the left-hand side is nonnegative, so the inequality holds automatically. Combined with the domain condition, it is satisfied for:
\[ -1\le x\le 1. \]
Case 2: \(x-1\gt 0\)
If:
\[ x-1\gt 0, \]
then \(x\gt 1\). Both sides are nonnegative, so we may square:
\[ x+1\ge (x-1)^2. \]
Expanding:
\[ x+1\ge x^2-2x+1. \]
Rearranging:
\[ x^2-3x\le 0. \]
Factoring:
\[ x(x-3)\le 0, \]
which gives:
\[ 0\le x\le 3. \]
Intersecting with \(x\gt 1\):
\[ 1\lt x\le 3. \]
Taking the union of both cases:
\[ [-1,1]\cup(1,3]=[-1,3]. \]
Therefore:
\[ S=[-1,3]. \]
Exercise 5 — level ★★☆☆☆
Solve:
\[ \sqrt{x-2}\lt x-4. \]
Answer
\[ S=(6,+\infty). \]
Solution
The radical is defined when:
\[ x-2\ge 0, \]
that is:
\[ x\ge 2. \]
Since the left-hand side is nonnegative, for the inequality to hold we also need:
\[ x-4\gt 0, \]
that is:
\[ x\gt 4. \]
Now both sides are positive, so we may square:
\[ x-2\lt (x-4)^2. \]
Expanding:
\[ x-2\lt x^2-8x+16. \]
Rearranging:
\[ x^2-9x+18\gt 0. \]
Factoring:
\[ x^2-9x+18=(x-3)(x-6). \]
The inequality \((x-3)(x-6)\gt 0\) holds when:
\[ x\lt 3 \quad \text{or} \quad x\gt 6. \]
Intersecting with \(x\gt 4\):
\[ S=(6,+\infty). \]
Exercise 6 — level ★★☆☆☆
Solve:
\[ \sqrt{2x-1}\le x+1. \]
Answer
\[ S=\left[\frac{1}{2},+\infty\right). \]
Solution
The domain condition is:
\[ 2x-1\ge 0, \]
that is:
\[ x\ge \frac{1}{2}. \]
On this domain, \(x+1\gt 0\), so we may square both sides:
\[ 2x-1\le (x+1)^2. \]
Expanding:
\[ 2x-1\le x^2+2x+1. \]
Subtracting \(2x-1\) from both sides:
\[ 0\le x^2+2. \]
This is always true. The solution set is therefore just the domain:
\[ S=\left[\frac{1}{2},+\infty\right). \]
Exercise 7 — level ★★☆☆☆
Solve:
\[ \sqrt{x+3}\ge 2x-1. \]
Answer
\[ S=\left[-3,\frac{5+\sqrt{57}}{8}\right]. \]
Solution
The radical is defined when:
\[ x+3\ge 0, \]
that is:
\[ x\ge -3. \]
If \(2x-1\le 0\), i.e., \(x\le \frac{1}{2}\), the inequality holds automatically on the domain, giving:
\[ -3\le x\le \frac{1}{2}. \]
If instead \(2x-1\gt 0\), i.e., \(x\gt \frac{1}{2}\), both sides are positive and we may square:
\[ x+3\ge (2x-1)^2. \]
Expanding:
\[ x+3\ge 4x^2-4x+1. \]
Rearranging:
\[ 4x^2-5x-2\le 0. \]
Solving the associated equation \(4x^2-5x-2=0\), the discriminant is:
\[ \Delta=57. \]
The roots are:
\[ x=\frac{5-\sqrt{57}}{8} \quad \text{and} \quad x=\frac{5+\sqrt{57}}{8}. \]
So the quadratic inequality holds for:
\[ \frac{5-\sqrt{57}}{8}\le x\le \frac{5+\sqrt{57}}{8}. \]
Intersecting with \(x\gt \frac{1}{2}\):
\[ \frac{1}{2}\lt x\le \frac{5+\sqrt{57}}{8}. \]
Taking the union of both cases:
\[ S=\left[-3,\frac{5+\sqrt{57}}{8}\right]. \]
Exercise 8 — level ★★☆☆☆
Solve:
\[ \sqrt{x+5}\gt \sqrt{2x-1}. \]
Answer
\[ S=\left[\frac{1}{2},6\right). \]
Solution
The domain conditions are:
\[ \begin{cases} x+5\ge 0,\\ 2x-1\ge 0, \end{cases} \]
which together give:
\[ x\ge \frac{1}{2}. \]
Both sides are square roots (hence nonnegative), so we may square:
\[ x+5\gt 2x-1. \]
Hence:
\[ x\lt 6. \]
Intersecting with the domain:
\[ S=\left[\frac{1}{2},6\right). \]
Exercise 9 — level ★★★☆☆
Solve:
\[ \sqrt{x^2-1}\le 2. \]
Answer
\[ S=[-\sqrt{5},-1]\cup[1,\sqrt{5}]. \]
Solution
The radical is defined when:
\[ x^2-1\ge 0, \]
that is:
\[ x\le -1 \quad \text{or} \quad x\ge 1. \]
Since the right-hand side is positive, we may square:
\[ x^2-1\le 4, \]
so:
\[ x^2\le 5, \]
which gives:
\[ -\sqrt{5}\le x\le \sqrt{5}. \]
Intersecting with the domain:
\[ S=[-\sqrt{5},-1]\cup[1,\sqrt{5}]. \]
Exercise 10 — level ★★★☆☆
Solve:
\[ \sqrt{x^2-4x}\ge x-2. \]
Answer
\[ S=(-\infty,0]. \]
Solution
The radical is defined when:
\[ x^2-4x\ge 0. \]
Factoring:
\[ x(x-4)\ge 0, \]
which holds for:
\[ x\le 0 \quad \text{or} \quad x\ge 4. \]
If \(x\le 0\), then \(x-2\lt 0\), while the left-hand side is nonnegative. The inequality therefore holds for all such values.
If \(x\ge 4\), both sides are nonnegative and we may square:
\[ x^2-4x\ge (x-2)^2. \]
Expanding:
\[ x^2-4x\ge x^2-4x+4. \]
This simplifies to \(0\ge 4\), which is false. No solutions arise from this case.
Therefore:
\[ S=(-\infty,0]. \]
Exercise 11 — level ★★★☆☆
Solve:
\[ \sqrt{x^2-4x}\le x-2. \]
Answer
\[ S=[4,+\infty). \]
Solution
From the domain condition:
\[ x^2-4x\ge 0, \]
we get:
\[ x\le 0 \quad \text{or} \quad x\ge 4. \]
Moreover, since the left-hand side is nonnegative, we also need:
\[ x-2\ge 0, \]
that is, \(x\ge 2\). Intersecting, we are left with:
\[ x\ge 4. \]
Squaring both sides:
\[ x^2-4x\le (x-2)^2=x^2-4x+4. \]
This simplifies to \(0\le 4\), which is always true. Therefore:
\[ S=[4,+\infty). \]
Exercise 12 — level ★★★☆☆
Solve:
\[ \sqrt{3x+1}\gt x+1. \]
Answer
\[ S=(0,1). \]
Solution
The domain condition is:
\[ 3x+1\ge 0, \]
that is:
\[ x\ge -\frac{1}{3}. \]
On this domain, \(x+1\gt 0\), so we may square both sides:
\[ 3x+1\gt (x+1)^2. \]
Expanding:
\[ 3x+1\gt x^2+2x+1. \]
Rearranging:
\[ x^2-x\lt 0. \]
Factoring:
\[ x(x-1)\lt 0, \]
which holds for:
\[ 0\lt x\lt 1. \]
Therefore:
\[ S=(0,1). \]
Exercise 13 — level ★★★★☆
Solve:
\[ \sqrt{x+2}+1\gt x. \]
Answer
\[ S=\left[-2,\frac{3+\sqrt{13}}{2}\right). \]
Solution
The radical is defined when:
\[ x+2\ge 0, \]
that is:
\[ x\ge -2. \]
Moving the constant to the right-hand side:
\[ \sqrt{x+2}\gt x-1. \]
If \(x-1\lt 0\), i.e., \(x\lt 1\), the inequality holds automatically on the domain, giving:
\[ -2\le x\lt 1. \]
If \(x\ge 1\), both sides are nonnegative and we may square:
\[ x+2\gt (x-1)^2. \]
Expanding:
\[ x+2\gt x^2-2x+1. \]
Rearranging:
\[ x^2-3x-1\lt 0. \]
The roots of \(x^2-3x-1=0\) are:
\[ x=\frac{3-\sqrt{13}}{2} \quad \text{and} \quad x=\frac{3+\sqrt{13}}{2}. \]
The quadratic is negative between its roots:
\[ \frac{3-\sqrt{13}}{2}\lt x\lt \frac{3+\sqrt{13}}{2}. \]
Intersecting with \(x\ge 1\):
\[ 1\le x\lt \frac{3+\sqrt{13}}{2}. \]
Taking the union of both cases:
\[ S=\left[-2,\frac{3+\sqrt{13}}{2}\right). \]
Exercise 14 — level ★★★★☆
Solve:
\[ \sqrt{x+4}+\sqrt{x-1}\gt 5. \]
Answer
\[ S=(5,+\infty). \]
Solution
The domain conditions are:
\[ \begin{cases} x+4\ge 0,\\ x-1\ge 0, \end{cases} \]
which give:
\[ x\ge 1. \]
The function:
\[ f(x)=\sqrt{x+4}+\sqrt{x-1} \]
is strictly increasing on the domain. We therefore find the value of \(x\) at which \(f(x)=5\) by solving:
\[ \sqrt{x+4}+\sqrt{x-1}=5. \]
Isolating one radical:
\[ \sqrt{x+4}=5-\sqrt{x-1}. \]
Squaring both sides:
\[ x+4=25-10\sqrt{x-1}+x-1. \]
Simplifying:
\[ 10\sqrt{x-1}=20, \]
so:
\[ \sqrt{x-1}=2, \]
hence \(x=5\). Since \(f\) is strictly increasing and the inequality is strict:
\[ S=(5,+\infty). \]
Exercise 15 — level ★★★★☆
Solve:
\[ \sqrt{x+4}-\sqrt{x-1}\le 1. \]
Answer
\[ S=[5,+\infty). \]
Solution
The domain is:
\[ x\ge 1. \]
Moving the second radical to the right-hand side:
\[ \sqrt{x+4}\le 1+\sqrt{x-1}. \]
The right-hand side is positive, so we may square:
\[ x+4\le \left(1+\sqrt{x-1}\right)^2. \]
Expanding:
\[ x+4\le 1+2\sqrt{x-1}+(x-1)=x+2\sqrt{x-1}. \]
Subtracting \(x\) from both sides:
\[ 4\le 2\sqrt{x-1}, \]
so:
\[ 2\le \sqrt{x-1}. \]
Squaring:
\[ 4\le x-1, \]
hence \(x\ge 5\). Therefore:
\[ S=[5,+\infty). \]
Exercise 16 — level ★★★★☆
Solve:
\[ \sqrt{2x+3}\ge \sqrt{x+1}+1. \]
Answer
\[ S=\{-1\}\cup[3,+\infty). \]
Solution
The domain is:
\[ x\ge -1. \]
Both sides are nonnegative on the domain, so we may square:
\[ 2x+3\ge \left(\sqrt{x+1}+1\right)^2. \]
Expanding:
\[ 2x+3\ge (x+1)+2\sqrt{x+1}+1=x+2+2\sqrt{x+1}. \]
Simplifying:
\[ x+1\ge 2\sqrt{x+1}. \]
Setting \(t=\sqrt{x+1}\ge 0\), this becomes:
\[ t^2\ge 2t, \]
i.e.:
\[ t(t-2)\ge 0. \]
Since \(t\ge 0\), the inequality holds when \(t=0\) or \(t\ge 2\), which gives:
\[ x=-1 \quad \text{or} \quad x\ge 3. \]
Therefore:
\[ S=\{-1\}\cup[3,+\infty). \]
Exercise 17 — level ★★★★☆
Solve:
\[ \sqrt{x+6}\gt x. \]
Answer
\[ S=[-6,3). \]
Solution
The domain is:
\[ x\ge -6. \]
If \(x\lt 0\), the right-hand side is negative while the left-hand side is nonnegative, so the inequality holds for all:
\[ -6\le x\lt 0. \]
If \(x\ge 0\), both sides are nonnegative and we may square:
\[ x+6\gt x^2, \]
i.e.:
\[ x^2-x-6\lt 0. \]
Factoring:
\[ (x-3)(x+2)\lt 0, \]
which holds for \(-2\lt x\lt 3\). Intersecting with \(x\ge 0\):
\[ 0\le x\lt 3. \]
Taking the union of both cases:
\[ S=[-6,3). \]
Exercise 18 — level ★★★★★
Solve:
\[ \sqrt{x+1}+\sqrt{x+4}\le 5. \]
Answer
\[ S=\left[-1,\frac{96}{25}\right]. \]
Solution
The domain is:
\[ x\ge -1. \]
The function:
\[ f(x)=\sqrt{x+1}+\sqrt{x+4} \]
is strictly increasing on the domain. We find the boundary by solving \(f(x)=5\):
\[ \sqrt{x+1}+\sqrt{x+4}=5. \]
Isolating one radical:
\[ \sqrt{x+4}=5-\sqrt{x+1}. \]
Squaring:
\[ x+4=25-10\sqrt{x+1}+(x+1). \]
Simplifying:
\[ 10\sqrt{x+1}=22, \]
so:
\[ \sqrt{x+1}=\frac{11}{5}. \]
Squaring:
\[ x+1=\frac{121}{25}, \]
hence:
\[ x=\frac{96}{25}. \]
Since \(f\) is strictly increasing and the inequality is non-strict:
\[ S=\left[-1,\frac{96}{25}\right]. \]
Exercise 19 — level ★★★★★
Solve:
\[ \sqrt{x+2}\ge x. \]
Answer
\[ S=[-2,2]. \]
Solution
The domain is:
\[ x\ge -2. \]
If \(x\lt 0\), the right-hand side is negative and the inequality holds automatically for:
\[ -2\le x\lt 0. \]
If \(x\ge 0\), both sides are nonnegative and we may square:
\[ x+2\ge x^2, \]
i.e.:
\[ x^2-x-2\le 0. \]
Factoring:
\[ (x-2)(x+1)\le 0, \]
which holds for \(-1\le x\le 2\). Intersecting with \(x\ge 0\):
\[ 0\le x\le 2. \]
Taking the union of both cases:
\[ S=[-2,2]. \]
Exercise 20 — level ★★★★★
Solve:
\[ \sqrt{x+3}-\sqrt{x-1}\ge 1. \]
Answer
\[ S=\left[1,\frac{13}{4}\right]. \]
Solution
The domain is:
\[ x\ge 1. \]
Moving the second radical to the right-hand side:
\[ \sqrt{x+3}\ge 1+\sqrt{x-1}. \]
Both sides are nonnegative, so we may square:
\[ x+3\ge \left(1+\sqrt{x-1}\right)^2. \]
Expanding:
\[ x+3\ge 1+2\sqrt{x-1}+(x-1)=x+2\sqrt{x-1}. \]
Subtracting \(x\) from both sides:
\[ 3\ge 2\sqrt{x-1}, \]
so:
\[ \sqrt{x-1}\le \frac{3}{2}. \]
Squaring:
\[ x-1\le \frac{9}{4}, \]
hence \(x\le \frac{13}{4}\). Intersecting with the domain:
\[ S=\left[1,\frac{13}{4}\right]. \]