Solving Radical Equations

To solve an equation with the variable under a radical, isolate the radical and raise both sides to a matching power — but that step can invent an extraneous solution, so every candidate must be checked back in the original equation.

By the end you'll be able to isolate a radical, square both sides to solve, and tell a true solution from an extraneous one by substituting back.

Predict: if you drag k below 0, can squaring \(\sqrt{x+c}=k\) still hand you a "solution" — even though a square root can never equal a negative number? Drag the slider to check, then verify by substituting the candidate back in.

This is \(\sqrt{x+c} = k\), where c shifts the radicand and k is the target value. Squaring both sides always produces one candidate \(x = k^2 - c\), but that candidate only solves the original equation when \(k \ge 0\) — the true-solution point sits right on the curve. When \(k < 0\), the candidate lands where the curve reads \(|k|\), not \(k\) — the gap between the curve and the dashed line flags it as extraneous.

Squaring gives x = 3.0. Check: √(3.0 + 1.0) = 2.0, which equals k = 2.0 → true solution.

y = √(x + c) and y = k — where squaring's candidate really lands
y = √(x+c) y = k true solution extraneous candidate

Radical equations are solved by isolating a radical and raising both sides to a power, then checking every candidate for an extraneous solution introduced by that step.

Procedural

Follow the same four steps every time:

  • Isolate one radical on one side of the equation.
  • Raise both sides to the index of that radical — square a square root, cube a cube root.
  • Solve the resulting equation. If a radical remains, repeat isolate-and-raise.
  • Check every candidate in the original equation and discard any that fail.

Step 4 is never optional — it's the only reliable way to catch an extraneous root.

Formal

Squaring both sides of \(\sqrt{f(x)} = g(x)\) gives \(f(x) = g(x)^2\), but squaring erases sign: both \(g(x)\) and \(-g(x)\) square to the same thing, so the new equation can be satisfied by an x where \(g(x) < 0\) — impossible for the original equation, since \(\sqrt{f(x)} \ge 0\) always. That's exactly the gap you saw in the diagram: the curve and the k-line only truly meet when \(k \ge 0\).

Applied

This shows up whenever a model is built from a square root — solving for the time it takes a dropped object to fall a given distance, or the radius that gives a cylinder a target surface area. The algebra can hand you a number that's mathematically tidy but physically or logically impossible (a negative time, a radius that doesn't match the original shape), and checking against the original equation is what catches it.

Worked example

Solve \(\sqrt{2x-1} = x-2\). Already isolated. Square both sides: \(2x-1 = (x-2)^2 = x^2-4x+4\). Rearrange: \(0 = x^2-6x+5 = (x-1)(x-5)\), so the candidates are \(x=1\) and \(x=5\). Check \(x=1\): \(\sqrt{2(1)-1}=\sqrt{1}=1\), but \(x-2=-1\); since \(1 \ne -1\), reject it as extraneous. Check \(x=5\): \(\sqrt{2(5)-1}=\sqrt{9}=3\), and \(x-2=3\) — it checks. The only solution is x = 5.

Your turn

Solve \(\sqrt{3x+1} = x-3\). Already isolated. Square both sides: \(3x+1 = (x-3)^2 = x^2-6x+9\). Rearrange: \(0 = x^2-9x+8 = (x-1)(x-8)\), so the candidates are \(x=1\) and \(x=8\). Check \(x=1\): \(\sqrt{3(1)+1}=\sqrt{4}=2\), but \(x-3=-2\); since \(2 \ne -2\), this candidate is ____ — reject it. Now check \(x=8\): \(\sqrt{3(8)+1} = \sqrt{25}=\) ____, and \(x-3=\) ____.

Reveal the answer

\(x=1\) is extraneous — the right side is negative, which a square root can never equal. At \(x=8\): \(\sqrt{25}=\) 5, and \(x-3=\) 5 — they match, so the only solution is x = 8. Set c and k on the sliders above so that \(x+c=25\) and \(k=5\) (try c = 0, drag k to 5) and check the true-solution point lands exactly where you'd expect.

More info — why squaring can invent a solution

Squaring is not a reversible step the way adding or multiplying by a nonzero number is: \(a=b\) implies \(a^2=b^2\), but \(a^2=b^2\) only implies \(a=b\) OR \(a=-b\). When you square \(\sqrt{f(x)}=g(x)\), you're really solving the combined statement "\(g(x)=\sqrt{f(x)}\) or \(g(x)=-\sqrt{f(x)}\)" — and the original equation only ever meant the first half. Any candidate landing in the second half is extraneous. With two radicals (isolate one, square, isolate the remaining radical, square again) the same risk compounds at each squaring step — see the Purplemath link in Dive deeper for a worked two-radical example.

Check your understanding

Question 1 of 4

Solve \(\sqrt{x+5} = 3\) for x, and confirm whether it's a true solution.

Question 2 of 4

What is the solution set of \(\sqrt{x-2} = -3\)?

Question 3 of 4

Solve \(\sqrt{x+7} = x+1\), checking both candidates against the original equation.

Question 4 of 4

Solving \(\sqrt{x+7}=x+1\) means squaring both sides; solving \(\dfrac{x}{x-3}=\dfrac{3}{x-3}+2\) means clearing denominators by multiplying by \((x-3)\). Both moves can manufacture a candidate that fails in the original equation. What actually causes each one's extraneous risk?

Recap

  • Isolate one radical, then raise both sides to the index of that radical (square a square root, cube a cube root).
  • Solve the resulting equation; repeat isolate-and-raise if a radical remains.
  • Always check every candidate in the original equation — squaring erases sign information and can manufacture a candidate that satisfies the squared equation but not the original one.
  • A candidate is extraneous when it makes one side of the original equation negative where a principal root can only be \(\ge 0\).

Dive deeper

Sources

  • Solving Radical Equations