A practical guide to solving equations involving square and cube roots. Learn how to handle domain restrictions, isolate radicals, raise both sides to a power, and perform final checks to discard extraneous solutions.
Exercise 1 — level ★★☆☆☆
\[ \sqrt{x} = 5 \]
Result
\[ x = 25 \]
Solution
Domain restrictions
The radicand must be nonnegative: \(x \geq 0\).
Squaring both sides
The right-hand side is positive, so squaring preserves equivalence.
\[ x = 25 \]
Check
\[ \sqrt{25} = 5 \]
Final answer
\[ \boxed{x = 25} \]
Exercise 2 — level ★★☆☆☆
\[ \sqrt{x - 3} = 4 \]
Result
\[ x = 19 \]
Solution
Domain restrictions
\[ x - 3 \geq 0 \implies x \geq 3 \]
Squaring both sides
\[ x - 3 = 16 \implies x = 19 \]
Check
\[ \sqrt{19-3} = \sqrt{16} = 4 \]
Final answer
\[ \boxed{x = 19} \]
Exercise 3 — level ★★☆☆☆
\[ \sqrt{2x + 1} = 3 \]
Result
\[ x = 4 \]
Solution
Domain restrictions
\[ 2x + 1 \geq 0 \implies x \geq -\tfrac{1}{2} \]
Squaring both sides
\[ 2x + 1 = 9 \implies x = 4 \]
Check
\[ \sqrt{9} = 3 \]
Final answer
\[ \boxed{x = 4} \]
Exercise 4 — level ★★☆☆☆
\[ \sqrt{3x - 2} = \sqrt{x + 6} \]
Result
\[ x = 4 \]
Solution
Domain restrictions
\[ 3x-2 \geq 0 \implies x \geq \tfrac{2}{3} \qquad x+6 \geq 0 \implies x \geq -6 \]
Overall condition: \(x \geq \tfrac{2}{3}\).
Key idea
If two square roots are equal and both radicands are nonnegative, then the radicands themselves can be set equal.
Equating radicands
\[ 3x - 2 = x + 6 \implies 2x = 8 \implies x = 4 \]
Check
\[ \sqrt{10} = \sqrt{10} \]
Final answer
\[ \boxed{x = 4} \]
Exercise 5 — level ★★★☆☆
\[ \sqrt{x + 4} = x - 2 \]
Result
\[ x = 5 \]
Solution
Domain restrictions
\[ x+4 \geq 0 \implies x \geq -4 \qquad x-2 \geq 0 \implies x \geq 2 \]
The right-hand side must be nonnegative since it equals a square root. Dominant condition: \(x \geq 2\).
Squaring both sides
\[ x+4 = (x-2)^2 = x^2-4x+4 \implies x^2-5x = 0 \implies x(x-5) = 0 \]
Checking solutions
\(x = 0\): does not satisfy \(x \geq 2\), reject.
\(x = 5\): \(\sqrt{9} = 3 = 5-2\) — valid.
Final answer
\[ \boxed{x = 5} \]
Exercise 20 — level ★★★★★
\[ \sqrt[3]{x + 1} + \sqrt[3]{x - 1} = \sqrt[3]{2x} \]
Result
\[ x = 0 \quad \text{or} \quad x = 1 \quad \text{or} \quad x = -1 \]
Solution
Strategy
Use the identity \((a+b)^3 = a^3+b^3+3ab(a+b)\), with \(a=\sqrt[3]{x+1}\), \(b=\sqrt[3]{x-1}\), and \(a+b=\sqrt[3]{2x}\).
Cubing both sides
\[ (x+1)+(x-1)+3\sqrt[3]{(x+1)(x-1)}\cdot\sqrt[3]{2x} = 2x \]
\[ 2x + 3\sqrt[3]{2x(x^2-1)} = 2x \implies 3\sqrt[3]{2x(x^2-1)} = 0 \]
Solution
\[ 2x(x^2-1) = 0 \implies x = 0,\; x = 1,\; x = -1 \]
Check
All values satisfy the original equation.
Final answer
\[ \boxed{x = 0 \quad \text{or} \quad x = 1 \quad \text{or} \quad x = -1} \]