Solving Radical Equations: Practice

The radical already stands alone, so square both sides. Squaring undoes the square root on the left.

$\left(\sqrt{x}\right)^2 = 7^2 \;\Longrightarrow\; x = 49$

Check the candidate in the original equation: $\sqrt{49} = 7$, which is correct. So $x = 49$.

Squaring both sides gives a candidate, but a candidate is not yet a solution.

$x = (-5)^2 = 25$

Check $x = 25$ in the original. The left side is $\sqrt{25} = 5$, and the right side is $-5$, so the equation reads $5 = -5$, which is false.

The principal square root is never negative, so no value of $x$ can make $\sqrt{x}$ equal $-5$. The candidate is extraneous and the equation has no solution.

Square both sides to remove the radical.

$2x + 1 = 25$

Subtract $1$ and divide by $2$: $2x = 24$, so $x = 12$.

Check in the original: $\sqrt{2(12) + 1} = \sqrt{25} = 5$, which is correct. So $x = 12$.

Work inside the radical first, because the radical sign groups everything beneath it.

$\sqrt{(-6)^2} = \sqrt{36} = 6$

The answer is $6$, not $-6$. This is the identity $\sqrt{t^2} = \lvert t \rvert$ in action: squaring destroyed the minus sign, and the principal root cannot put it back.

A cube root is allowed to be negative, unlike a principal square root, so a negative right side is no obstacle here. Cube both sides.

$x = (-2)^3 = -8$

Check in the original: $\sqrt[3]{-8} = -2$, since $(-2)^3 = -8$. So $x = -8$. Cubing is reversible, so no extraneous root can appear.

Substitute $x = 4$ into the original equation and evaluate each side separately.

$\text{left} = \sqrt{4 + 5} = \sqrt{9} = 3, \qquad \text{right} = 4 - 1 = 3$

The two sides agree, so $x = 4$ is a solution. The radicand $9$ is positive, so $x = 4$ is comfortably inside the domain as well.

Square both sides to clear the radical.

$4x - 3 = 9 \;\Longrightarrow\; 4x = 12 \;\Longrightarrow\; x = 3$

Check in the original: $\sqrt{4(3) - 3} = \sqrt{9} = 3$, which is correct. So $x = 3$.

Isolate the radical before raising anything to a power. Subtracting $3$ leaves the radical alone on the left.

$\sqrt{x + 1} = 4$

Squaring first would not help, because squaring a sum keeps a cross term: $\left(\sqrt{x+1} + 3\right)^2 = (x + 1) + 6\sqrt{x + 1} + 9$, and the radical survives. Subtracting $1$ from both sides does nothing useful either, since the $1$ is inside the radical and cannot be pulled out of it.

The squaring itself was done correctly: $(-4)^2 = 16$. The problem is that squaring is not reversible, so the candidate must be tested in the original equation.

$\sqrt{16} = 4 \ne -4$

The candidate fails, so it is extraneous and the equation has no solution. Note that $x = 16$ is inside the domain of $\sqrt{x}$, since $16$ is non-negative. It fails on sign, not on domain, which is exactly why a domain check would have missed it.

A rational exponent of $\tfrac{1}{2}$ is another name for a principal square root, so this is the equation $\sqrt{x} = 6$ in different clothes. Raise both sides to the power $2$.

$\left(x^{1/2}\right)^2 = x^1 = 6^2 = 36$

Check in the original: $36^{1/2} = \sqrt{36} = 6$, which is correct. So $x = 36$.

Isolate the radical first, and read what it says before doing anything else.

$\sqrt{x} = 1 - 3 = -2$

A principal square root is never negative, so no value of $x$ can satisfy this. The equation has no solution.

If you square anyway you get the candidate $x = 4$, but checking it in the original gives $\sqrt{4} + 3 = 2 + 3 = 5 \ne 1$, so it is extraneous.

Isolate the radical by adding $1$ to both sides.

$\sqrt{2x} = 4$

Now square both sides: $2x = 16$, so $x = 8$.

Check in the original: $\sqrt{2(8)} - 1 = \sqrt{16} - 1 = 4 - 1 = 3$, which is correct. So $x = 8$.