Solving Radical Equations

🌱 Solving Radical Equations

Part 1 of 5 — Square Roots & the Big Idea

Topics in This Part

🔑 Key Concept: A radical equation has the variable trapped underneath a root sign. To free it, you undo the root with its inverse — squaring undoes a square root. Everything in this lesson grows from that one idea.

What Is a Radical Equation?

A radical equation is any equation where the variable lives inside a radical (a root):

$\sqrt{x} = 5, \qquad \sqrt{x + 3} = 4, \qquad \sqrt{2x - 1} = 3$

The expression under the root sign is called the radicand. Our whole job is to get the variable out from under that root.

The Inverse of a Square Root

Squaring and square-rooting are inverse operations — each undoes the other:

$\left(\sqrt{x}\right)^2 = x \qquad\text{and}\qquad \sqrt{x^2} = |x|$

So to solve $\sqrt{x} = 5$, square both sides:

$\left(\sqrt{x}\right)^2 = 5^2 \;\Rightarrow\; x = 25$

✅ Check: $\sqrt{25} = 5$ ✓

Concept Check 🎯

Square to Solve 🧮

Solve each equation by squaring both sides.

1) $\sqrt{x} = 7,\quad x = \,?$ 2) $\sqrt{x} = 12,\quad x = \,?$ 3) $\sqrt{x + 3} = 4,\quad x = \,?$

Isolate the Radical First

Squaring only cleanly removes the root when the radical is alone on one side. If something is added, subtracted, or multiplied outside the root, deal with that first.

Worked Example: $\sqrt{x} + 2 = 9$

If you squared right now, the left side would become a messy $\left(\sqrt{x}+2\right)^2$. Instead, isolate the radical:

$\sqrt{x} + 2 = 9 \;\Rightarrow\; \sqrt{x} = 7 \;\Rightarrow\; x = 49$

Worked Example: $3\sqrt{x} = 12$

Divide first, then square:

$3\sqrt{x} = 12 \;\Rightarrow\; \sqrt{x} = 4 \;\Rightarrow\; x = 16$

⚠️ Golden Rule: Isolate the radical before you square. Squaring a sum like $\sqrt{x}+2$ creates a brand-new radical and makes the problem harder, not easier.

Isolate, Then Square 🔽

You're solving $2\sqrt{x} - 1 = 9$. Choose what happens at each stage.

🌱 Solving Radical Equations

Part 2 of 5 — Extraneous Solutions

🔑 The Catch: Squaring both sides can create solutions that don't actually work. These fake answers are called extraneous solutions. You must check every answer in the original equation.

Why Fake Solutions Appear

Squaring is not perfectly reversible. Look:

$3 = 3 \quad\text{is true, and}\quad (-3)^2 = 3^2 \quad\text{is also true.}$

Squaring erases the sign of a number. So when you square both sides of an equation, you may accidentally pull in a "negative-version" answer that the original equation never allowed.

The Tell-Tale Sign

The principal square root symbol $\sqrt{\;}$ only ever outputs a non-negative number. So:

$\sqrt{x} = -2 \quad\text{has NO solution.}$

But if you square it carelessly, you get $x = 4$ — and $\sqrt{4} = 2 \ne -2$. That $x=4$ is extraneous.

⚠️ Mandatory Step: After solving, substitute each candidate back into the original equation. Keep only the ones that make it true.

Concept Check 🎯

Worked Example: Catching an Extraneous Root

Solve $\sqrt{x + 2} = x$.

Step 1 — Square both sides: $x + 2 = x^2$

Step 2 — Rearrange into a quadratic: $x^2 - x - 2 = 0 \;\Rightarrow\; (x - 2)(x + 1) = 0$

Step 3 — Candidates: $x = 2$ or $x = -1$.

Step 4 — Check BOTH in the original:

The only real solution is $\boxed{x = 2}$. The candidate $x=-1$ fails because $\sqrt{1}=1$, not $-1$.

Valid or Extraneous? 🔽

Solve $\sqrt{x + 7} = x + 1$. Squaring gives $x^2 + x - 6 = 0$, so the candidates are $x = 2$ and $x = -3$. Check each.

Solve and Verify 🧮

Solve each, then keep only the valid root(s). Enter the valid solution.

1) $\sqrt{x} = -3 \Rightarrow$ how many solutions? (enter the number 0, 1, or 2) 2) $\sqrt{x - 4} = x - 6 \Rightarrow$ the single valid solution is $x = \,?$

🌱 Solving Radical Equations

Part 3 of 5 — Equations That Become Quadratics

🔑 Pattern: When you square a radical equal to a binomial (like $x-2$), the squared side becomes a quadratic. Solve it by factoring or the quadratic formula, then check for extraneous roots.

Squaring a Binomial Correctly

The number-one error here is forgetting the middle term. Remember:

$(x - 3)^2 = x^2 - 6x + 9 \quad\text{NOT}\quad x^2 - 9$

Use the rule $(a \pm b)^2 = a^2 \pm 2ab + b^2$ every single time.

Worked Example: $\sqrt{x + 5} = x - 1$

Square both sides (watch the right side!): $x + 5 = (x - 1)^2 = x^2 - 2x + 1$

Move everything to one side: $0 = x^2 - 3x - 4 = (x - 4)(x + 1)$

Candidates: $x = 4$ or $x = -1$.

Check:

• $x = 4$: $\sqrt{9} = 3$ and $4 - 1 = 3$ ✓
• $x = -1$: $\sqrt{4} = 2$ but $-1 - 1 = -2$ ✗ (extraneous)

Solution: $x = 4$.

Concept Check 🎯

Worked Example: Both Candidates Valid

Solve $\sqrt{3x + 1} = x + 1$.

Square both sides: $3x + 1 = (x + 1)^2 = x^2 + 2x + 1$

Simplify: $0 = x^2 - x = x(x - 1)$

Candidates: $x = 0$ or $x = 1$.

Check:

• $x = 0$: $\sqrt{1} = 1$ and $0 + 1 = 1$ ✓
• $x = 1$: $\sqrt{4} = 2$ and $1 + 1 = 2$ ✓

💡 Both answers work! Extraneous roots are common but not guaranteed. That's exactly why you check — to find out which (if any) survive. Solution: $x = 0$ and $x = 1$.

Build the Quadratic 🔽

You're solving $\sqrt{x + 3} = x - 3$. Choose each step.

Solve the Quadratic-Producing Equation 🧮

Solve each and enter the valid solution.

1) $\sqrt{x + 6} = x \Rightarrow x = \,?$ (one valid root) 2) $\sqrt{5x - 1} = x + 1 \Rightarrow$ enter the larger valid root, $x = \,?$

🌱 Solving Radical Equations

Part 4 of 5 — Cube Roots & Two Radicals

🔑 Two New Tools: (1) Cube roots are undone by cubing and never produce extraneous solutions. (2) When two radicals appear, isolate one, square, and sometimes square a second time.

Cube Roots (and Higher Odd Roots)

A cube root is undone by cubing both sides:

$\left(\sqrt[3]{x}\right)^3 = x$

Worked Example: $\sqrt[3]{2x - 3} = 3$

Cube both sides: $2x - 3 = 3^3 = 27 \;\Rightarrow\; 2x = 30 \;\Rightarrow\; x = 15$

✅ Check: $\sqrt[3]{2(15) - 3} = \sqrt[3]{27} = 3$ ✓

Why No Extraneous Solutions?

Cube roots accept and return negative numbers: $\sqrt[3]{-8} = -2$. Cubing does not erase sign information the way squaring does, so odd roots never create extraneous solutions. You still should check for arithmetic slips, but the sign trap is gone.

💡 Rule of thumb: Even roots (square, fourth, …) can spawn extraneous solutions; odd roots (cube, fifth, …) cannot.

Concept Check 🎯

Cube-Root Practice 🧮

Solve each by cubing both sides.

1) $\sqrt[3]{x} = 2 \Rightarrow x = \,?$ 2) $\sqrt[3]{x - 1} = 3 \Rightarrow x = \,?$ 3) $\sqrt[3]{x} = -5 \Rightarrow x = \,?$ (odd roots allow negatives!)

Two Radicals

Case A: A Radical on Each Side

If both sides are a single radical, square once and the roots vanish together.

Solve $\sqrt{x + 5} = \sqrt{2x - 1}$: $x + 5 = 2x - 1 \;\Rightarrow\; 6 = x$

✅ Check: $\sqrt{6+5} = \sqrt{11}$ and $\sqrt{2(6)-1} = \sqrt{11}$ ✓

Case B: A Radical Plus a Term

When a loose term is in the way, you may need to square twice.

Solve $\sqrt{x + 4} = x - 8$ (already isolated): $x + 4 = (x-8)^2 = x^2 - 16x + 64 \;\Rightarrow\; x^2 - 17x + 60 = 0$ $(x - 5)(x - 12) = 0 \;\Rightarrow\; x = 5 \text{ or } 12$

Check: $x=12$: $\sqrt{16}=4=12-8$ ✓. $\;x=5$: $\sqrt{9}=3$ but $5-8=-3$ ✗. Solution $x=12$.

Two Radicals, One Square 🔽

Solve $\sqrt{3x + 4} = \sqrt{x + 10}$.

🌱 Solving Radical Equations

Part 5 of 5 — Applications & Mastery Check

You can now (1) isolate and square, (2) catch extraneous solutions, (3) solve equations that become quadratics, and (4) handle cube roots and two radicals. Time to apply it and prove mastery.

Quick Reference

⚠️ Never skip the check for even roots. The principal square root is always $\ge 0$, so any "solution" making a root equal a negative is extraneous.

Application: A Real Formula

The time $t$ (in seconds) for an object to fall a distance $d$ (in feet) follows:

$t = \frac{1}{4}\sqrt{d}$

Question: How far does an object fall in $t = 3$ seconds?

Solve for $d$: $3 = \frac{1}{4}\sqrt{d} \;\Rightarrow\; 12 = \sqrt{d} \;\Rightarrow\; d = 144 \text{ feet}$

✅ Check: $\frac{1}{4}\sqrt{144} = \frac{1}{4}(12) = 3$ ✓

Radical equations model falling objects, pendulum periods, and electrical formulas — isolating the radical and squaring is the same skill every time.

Apply It 🧮

Using the pendulum-style formula $T = 2\sqrt{L}$ (period $T$ in seconds, length $L$ in meters):

1) If $T = 6$, isolate the radical: $\sqrt{L} = \,?$ 2) Solve for the length $L = \,?$ meters.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.