Solving Radical Equations: Practice

Isolate one radical and square both sides.

$\sqrt{x + 4} = 5 - \sqrt{x - 1} \;\Longrightarrow\; x + 4 = 25 - 10\sqrt{x - 1} + (x - 1)$

The $x$ cancels, leaving $10\sqrt{x - 1} = 20$, so $\sqrt{x - 1} = 2$ and $x - 1 = 4$, giving $x = 5$.

Check in the original: $\sqrt{9} + \sqrt{4} = 3 + 2 = 5$, which is correct. So $x = 5$.

Isolate one radical and square.

$\sqrt{2x - 1} = 4 - \sqrt{x - 4} \;\Longrightarrow\; 2x - 1 = 16 - 8\sqrt{x - 4} + (x - 4)$

Simplify to $x - 13 = -8\sqrt{x - 4}$, that is $13 - x = 8\sqrt{x - 4}$. Square again.

$169 - 26x + x^2 = 64x - 256 \;\Longrightarrow\; x^2 - 90x + 425 = 0 \;\Longrightarrow\; (x - 5)(x - 85) = 0$

Check $x = 5$: $\sqrt{9} + \sqrt{1} = 3 + 1 = 4$, which is correct.

Check $x = 85$: $\sqrt{169} + \sqrt{81} = 13 + 9 = 22 \ne 4$, so it is extraneous. It entered when we squared $13 - x = 8\sqrt{x-4}$, a step that ignores the fact that the left side must be non-negative. Only $x = 5$ solves the equation.

This is a quadratic in $\sqrt{x}$. Let $u = \sqrt{x}$, so $x = u^2$ and $u \ge 0$.

$2u^2 + u - 1 = 0 \;\Longrightarrow\; (2u - 1)(u + 1) = 0$

So $u = \tfrac{1}{2}$ or $u = -1$. Discard $u = -1$, since a principal square root is never negative. From $\sqrt{x} = \tfrac{1}{2}$ we get $x = \tfrac{1}{4}$.

Check in the original: $2\left(\tfrac{1}{4}\right) + \tfrac{1}{2} - 1 = \tfrac{1}{2} + \tfrac{1}{2} - 1 = 0$, which is correct.

Square both sides and solve the quadratic that appears.

$x + 5 = x^2 - 2x + 1 \;\Longrightarrow\; x^2 - 3x - 4 = 0 \;\Longrightarrow\; (x - 4)(x + 1) = 0$

The candidates are $x = 4$ and $x = -1$, but the count of candidates is not the count of solutions.

Check $x = 4$: the left side is $\sqrt{9} = 3$ and the right side is $3$, so it survives. Check $x = -1$: the left side is $\sqrt{4} = 2$ and the right side is $-2$, so it is extraneous. Exactly one candidate survives, so the equation has exactly one solution.

Cubing both sides is reversible, so the cubed equation says exactly what the original says.

$x^2 + 3x = 4x + 6 \;\Longrightarrow\; x^2 - x - 6 = 0 \;\Longrightarrow\; (x - 3)(x + 2) = 0$

Check $x = 3$: both radicands equal $18$, so both sides are $\sqrt[3]{18}$.

Check $x = -2$: both radicands equal $-2$, so both sides are $\sqrt[3]{-2}$, which is a perfectly good real number because a cube root accepts negative inputs. Both candidates are genuine solutions, and no check could have rejected either, since cubing manufactures no extraneous roots.

The exponent $\tfrac{1}{2}$ is a square root, so raise both sides to the power $2$.

$x^2 - 5x = 36 \;\Longrightarrow\; x^2 - 5x - 36 = 0 \;\Longrightarrow\; (x - 9)(x + 4) = 0$

Check $x = 9$: the radicand is $81 - 45 = 36$, so the left side is $\sqrt{36} = 6$, which matches.

Check $x = -4$: the radicand is $16 + 20 = 36$, so the left side is $\sqrt{36} = 6$, which also matches. The right side is the positive constant $6$, so squaring had no sign to destroy, and both candidates are solutions.

Isolate one radical and square.

$\sqrt{2x + 3} = 2 + \sqrt{x - 2} \;\Longrightarrow\; 2x + 3 = 4 + 4\sqrt{x - 2} + (x - 2)$

Simplify to $x + 1 = 4\sqrt{x - 2}$ and square again.

$x^2 + 2x + 1 = 16x - 32 \;\Longrightarrow\; x^2 - 14x + 33 = 0 \;\Longrightarrow\; (x - 3)(x - 11) = 0$

Check $x = 3$: $\sqrt{9} - \sqrt{1} = 3 - 1 = 2$, which is correct. Check $x = 11$: $\sqrt{25} - \sqrt{9} = 5 - 3 = 2$, which is also correct. Both are solutions.

The index is $4$, so raise both sides to the fourth power.

$x^2 - 9 = 2^4 = 16 \;\Longrightarrow\; x^2 = 25$

Undoing the square admits both signs, so the candidates are $x = 5$ and $x = -5$.

Check $x = 5$: the radicand is $25 - 9 = 16$ and $\sqrt[4]{16} = 2$, which is correct. Check $x = -5$: the radicand is again $16$, so the left side is again $2$. The unknown appears only as $x^2$, so its sign never reaches the equation, and both candidates are solutions.

Isolate one radical and square both sides.

$\sqrt{x + 16} = 8 - \sqrt{x} \;\Longrightarrow\; x + 16 = 64 - 16\sqrt{x} + x$

The $x$ cancels, leaving $16\sqrt{x} = 48$, so $\sqrt{x} = 3$ and $x = 9$.

Check in the original: $\sqrt{9} + \sqrt{25} = 3 + 5 = 8$, which is correct. So $x = 9$.

One radical is already isolated on the left, so square both sides and expand the binomial on the right.

$x + 2 = 4 + 4\sqrt{x - 6} + (x - 6) = x - 2 + 4\sqrt{x - 6}$

The $x$ cancels, leaving $4 = 4\sqrt{x - 6}$, so $\sqrt{x - 6} = 1$ and $x - 6 = 1$, giving $x = 7$.

Check in the original: the left side is $\sqrt{9} = 3$ and the right side is $2 + \sqrt{1} = 3$, which is correct. So $x = 7$.

The left side is a principal square root, so it can take any non-negative value and no negative value. If $k < 0$ there is nothing to do: no $x$ can work.

If $k \ge 0$, square both sides to produce a candidate and then verify it.

$x - 2 = k^2 \;\Longrightarrow\; x = k^2 + 2, \qquad \sqrt{(k^2 + 2) - 2} = \sqrt{k^2} = \lvert k \rvert = k$

The last equality uses $k \ge 0$, so the candidate genuinely works. The value $k = 0$ must be included, since $x = 2$ gives $\sqrt{0} = 0$, which is why $k > 0$ is too narrow.

Substitute each candidate into the original equation.

At $x = 3$ the left side is $\sqrt{4} = 2$ and the right side is $3 - 1 = 2$, so this candidate is genuine.

At $x = -2$ the left side is $\sqrt{9} = 3$ and the right side is $-2 - 1 = -3$:

$3 \ne -3$

So $x = -2$ is extraneous. Its radicand is $9$, which is positive, so it is inside the domain; it fails only because the right side turned negative while a principal square root cannot.