Radical equations
Radical equations have the variable inside a root. Isolating the radical and raising both sides to a power removes it, but this step can also introduce false answers.
Isolate, then power
To solve \(\sqrt{x + 3} = 5\), square both sides: \(x + 3 = 25\), so \(x = 22\). Check: \(\sqrt{25} = 5\). If the radical is not alone, isolate it first: \(\sqrt{2x - 1} + 4 = 7\) becomes \(\sqrt{2x - 1} = 3\), so \(2x - 1 = 9\) and \(x = 5\).
Why checking matters
Squaring both sides can create extraneous solutions, because \(a^2 = b^2\) does not mean \(a = b\). For \(\sqrt{x} = -4\), squaring gives x = 16, but \(\sqrt{16} = 4\), not -4. A principal square root is never negative, so there is no solution.
Equations leading to quadratics
Solve \(\sqrt{x + 6} = x\). Squaring: \(x + 6 = x^2\), so \(x^2 - x - 6 = 0\) and \((x - 3)(x + 2) = 0\). Checking: x = 3 gives \(\sqrt{9} = 3\), which works; x = -2 gives \(\sqrt{4} = 2 \neq -2\), which is extraneous. The only solution is x = 3.
Two radicals
With two radicals, isolate one and square, then isolate the remaining radical and square again. For \(\sqrt{x + 5} = 1 + \sqrt{x}\): squaring gives \(x + 5 = 1 + 2\sqrt{x} + x\), so \(2\sqrt{x} = 4\), \(\sqrt{x} = 2\), and x = 4. Check: \(\sqrt{9} = 3 = 1 + 2\).
Cube roots and other powers
For cube roots, cube both sides: \(\sqrt[3]{2x + 1} = 3\) gives \(2x + 1 = 27\), so x = 13. Odd powers do not create extraneous solutions the way squaring does, but checking is still a good habit. Equations with rational exponents work the same way: \(x^{2/3} = 9\) gives \(x = \pm 27\).
Radicals in formulas
Many formulas contain roots. A pendulum's period is \(T = 2\pi\sqrt{\frac{L}{g}}\). Solving for length: \(\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\), so \(L = g\left(\frac{T}{2\pi}\right)^2\). A 2-second pendulum on Earth, with g = 9.8 m/s², needs \(L \approx 9.8 \times 0.101 \approx 0.99\) m.
Graphical view
Solutions of \(\sqrt{x + 6} = x\) are the x-coordinates where the graphs of \(y = \sqrt{x + 6}\) and \(y = x\) meet. They meet only at x = 3; the extraneous answer -2 corresponds to where \(y = x\) meets the reflected curve \(y = -\sqrt{x + 6}\), which squaring sneaks in.
Worked example
The speed of a car before skidding can be estimated by \(s = \sqrt{30fd}\) in miles per hour, where d is the skid length in feet and f the road's friction factor. If a car skidded on a dry road with f = 0.75 and the estimated speed was 45 mph, then \(2025 = 22.5d\), so the skid was 90 feet long.
Key takeaways
Isolate the radical, then raise both sides to the matching power. Squaring can create extraneous solutions, so always check every answer in the original equation; a principal square root can never equal a negative number. Squaring may produce a quadratic, which you then solve and check. With two radicals, isolate and square twice. Cube both sides for cube roots. The same steps rearrange formulas like the pendulum period. Extraneous answers correspond to intersections with the reflected radical curve.
Flashcard
Variable inside a root
Check your understanding
Solve \(\sqrt{x + 6} = x\).