Basic Radical Equation

Solve: $\sqrt{x} = 6$

The radical is already isolated: $\sqrt{x} = 6$ = Ready to square

Square both sides: $(\sqrt{x})^2 = 6^2$ = $x = 36$

Check the solution: $\sqrt{36} = 6$ ✓ = Valid solution

Answer: $x = 36$

Radical with Expression Inside

Solve: $\sqrt{x + 5} = 4$

The radical is already isolated: $\sqrt{x + 5} = 4$ = Ready to square

Square both sides: $(\sqrt{x + 5})^2 = 4^2$ = $x + 5 = 16$

Solve for x: $x = 16 - 5$ = $x = 11$

Check the solution: $\sqrt{11 + 5} = \sqrt{16} = 4$ ✓ = Valid solution

Answer: $x = 11$

Isolating the Radical First

Solve: $\sqrt{2x + 3} - 5 = 0$

Isolate the radical: $\sqrt{2x + 3} = 5$ = Add 5 to both sides

Square both sides: $(\sqrt{2x + 3})^2 = 5^2$ = $2x + 3 = 25$

Solve for x: $2x = 22$, so $x = 11$ = $x = 11$

Check the solution: $\sqrt{2(11) + 3} - 5 = \sqrt{25} - 5 = 5 - 5 = 0$ ✓ = Valid solution

Answer: $x = 11$

Equation with Extraneous Solution

Solve: $\sqrt{x + 3} = x - 3$

Square both sides: $(\sqrt{x + 3})^2 = (x - 3)^2$ = $x + 3 = x^2 - 6x + 9$

Rearrange to standard form: $0 = x^2 - 7x + 6$ = Quadratic equation

Factor the quadratic: $0 = (x - 6)(x - 1)$ = $x = 6$ or $x = 1$

Check x = 6: $\sqrt{6 + 3} = \sqrt{9} = 3$ and $6 - 3 = 3$ ✓ = Valid

Check x = 1: $\sqrt{1 + 3} = \sqrt{4} = 2$ but $1 - 3 = -2$ ✗ = Extraneous!

Answer: $x = 6$ (only valid solution; $x = 1$ is extraneous)

No Valid Solutions

Solve: $\sqrt{x - 2} = -3$

Analyze the equation: The square root symbol $\sqrt{}$ represents the principal (positive) root = Left side is always $\geq 0$

Compare to right side: The right side is $-3$, which is negative = No positive number equals a negative

Conclusion: A square root cannot equal a negative number = No solution exists

Answer: No solution (the square root of a real number is never negative)

Mistake: Forgetting to check for extraneous solutions

Why: Squaring both sides can introduce false solutions. For example, in $\sqrt{x} = -3$, squaring gives $x = 9$, but $\sqrt{9} = 3 \neq -3$.

Correct: Always substitute your answer back into the ORIGINAL equation to verify.

Mistake: Squaring before isolating the radical

Why: If the radical is not alone, squaring creates a more complex equation. For $\sqrt{x} + 2 = 5$, squaring gives $x + 4\sqrt{x} + 4 = 25$, which is harder to solve.

Correct: First isolate: $\sqrt{x} = 3$, then square: $x = 9$.

Mistake: Only squaring the radical, not the entire side

Why: Both complete sides must be squared. $(\sqrt{x + 3})^2 \neq x + 3^2$

Correct: Square the entire expression: $(\sqrt{x + 3})^2 = x + 3$

Mistake: Assuming square root can be negative

Why: The principal square root is always non-negative. $\sqrt{25} = 5$, not $\pm 5$.

Correct: If $\sqrt{x} = -k$ where $k > 0$, there is no solution.

Pendulum Period

The period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$ where $L$ is the length and $g = 9.8$ m/s2 is gravity.

If a clock pendulum needs a period of 2 seconds, what length should it be? Solve $2 = 2\pi\sqrt{\frac{L}{9.8}}$

Free Fall Distance

The distance an object falls is related to time by $d = \frac{1}{2}gt^2$, or equivalently $t = \sqrt{\frac{2d}{g}}$.

How long does it take for a ball to fall 45 meters? Solve $t = \sqrt{\frac{2 \times 45}{9.8}}$

Q: Why do we get extraneous solutions?

A: Squaring is not a reversible operation - both $3^2$ and $(-3)^2$ equal 9. When we square both sides of an equation, we potentially introduce solutions from the negative case that don't satisfy the original.

Q: What if there are two radicals in the equation?

A: Isolate one radical, square both sides, then isolate the remaining radical and square again. You may need to square twice. Always check your solutions!

Q: Can a radical equation have no solution?

A: Yes! If the equation requires a square root to equal a negative number, there is no real solution. Also, all potential solutions might be extraneous.