Solving Radical Equations: Additional Practice (Set 1)

Square both sides and factor the quadratic.

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

Check $x = 5$: the left side is $\sqrt{25} = 5$ and the right side is $5$, so it survives.

Check $x = -1$: the left side is $\sqrt{1} = 1$ and the right side is $-1$, so it is extraneous. A principal square root cannot be negative, so any candidate that makes the right side negative is doomed. Only $x = 5$ solves the equation.

Isolate one radical and square both sides.

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

The $x$ cancels, leaving $8\sqrt{x - 4} = 8$, so $\sqrt{x - 4} = 1$ and $x - 4 = 1$, giving $x = 5$.

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

Isolate one radical and square.

$\sqrt{x} = 1 + \sqrt{x - 5} \;\Longrightarrow\; x = 1 + 2\sqrt{x - 5} + (x - 5)$

The $x$ cancels, leaving $4 = 2\sqrt{x - 5}$, so $\sqrt{x - 5} = 2$ and $x - 5 = 4$, giving $x = 9$.

Check in the original: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$, which is correct. So $x = 9$.

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

$3x - 2 = x^2 - 4x + 4 \;\Longrightarrow\; x^2 - 7x + 6 = 0 \;\Longrightarrow\; (x - 1)(x - 6) = 0$

Check $x = 6$: the left side is $(16)^{1/2} = 4$ and the right side is $4$, so it survives.

Check $x = 1$: the left side is $(1)^{1/2} = 1$ and the right side is $-1$, so it is extraneous. Only $x = 6$ solves the equation.

Split the exponent instead of guessing. Write $(x + 7)^{3/2} = \left[(x + 7)^{1/2}\right]^3$ and let $u = (x + 7)^{1/2}$, so that $u \ge 0$.

$u^3 = 27 \;\Longrightarrow\; u = 3$

Cubing is reversible, so $u = 3$ is the only possibility. Then $(x + 7)^{1/2} = 3$ gives $x + 7 = 9$, so $x = 2$.

Check in the original: $(2 + 7)^{3/2} = 9^{3/2} = \left(\sqrt{9}\right)^3 = 27$, which is correct.

Read the left side before computing anything. Write it as $\left[(x - 3)^{1/2}\right]^3$ and set $u = (x - 3)^{1/2}$, a principal square root, so $u \ge 0$ wherever it is defined.

$u \ge 0 \;\Longrightarrow\; u^3 \ge 0$

A non-negative number cubed is non-negative, so the left side can never equal $-8$. The equation has no solution.

The tempting candidate $x = 7$ fails, since $(7 - 3)^{3/2} = 4^{3/2} = 8$, which is $+8$ and not $-8$; and values like $x = -1$ do not even give a real left side.

Square both sides. Once again the $x^2$ terms cancel.

$x^2 - 5 = x^2 - 10x + 25 \;\Longrightarrow\; 10x = 30 \;\Longrightarrow\; x = 3$

Check the only candidate in the original. The left side is $\sqrt{9 - 5} = \sqrt{4} = 2$ and the right side is $3 - 5 = -2$.

$2 \ne -2$

The candidate is extraneous, so the equation has no solution. Its radicand is $4$, which is positive, so $x = 3$ is squarely inside the domain: this is a failure of sign, and a domain check would have missed it entirely.

The index $4$ is even, so this is a principal root and it is never negative, exactly like a square root. A cube root would have been fine here, but a fourth root is not.

Raise both sides to the fourth power to see what the algebra offers:

$x + 7 = (-2)^4 = 16 \;\Longrightarrow\; x = 9$

Check $x = 9$ in the original: $\sqrt[4]{16} = 2$, not $-2$. The candidate is extraneous, and the equation has no solution.

Substitute each candidate into the original equation.

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

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

$3 \ne -3$

So $x = -3$ is extraneous. It is inside the domain, since its radicand $9$ is positive; it is really a solution of the twin equation $\sqrt{6 - x} = -x$, which squaring folded into the problem.

Being in the domain only means both sides are defined; it says nothing about whether they are equal.

$\sqrt{x + 7} = x - 5 \;\text{ at } x = 2: \quad \sqrt{9} = 3, \quad 2 - 5 = -3$

The radicand is $9$, so $x = 2$ is inside the domain, yet the two sides come out $3$ and $-3$. Squaring cannot tell those apart, which is how the candidate arose in the first place. Isolating the radical first is good technique but changes nothing about this, so the only safe conclusion is that the candidate must be substituted into the original equation.

Isolate one radical and square both sides.

$\sqrt{x + 3} = 5 - \sqrt{x - 2} \;\Longrightarrow\; x + 3 = 25 - 10\sqrt{x - 2} + (x - 2)$

The $x$ cancels, leaving $10\sqrt{x - 2} = 20$, so $\sqrt{x - 2} = 2$ and $x - 2 = 4$, giving $x = 6$.

Check in the original: $\sqrt{9} + \sqrt{4} = 3 + 2 = 5$, which is correct. So $x = 6$.

Isolate one radical and square.

$\sqrt{5x + 1} = 1 + \sqrt{3x} \;\Longrightarrow\; 5x + 1 = 1 + 2\sqrt{3x} + 3x$

Simplify to $2x = 2\sqrt{3x}$, that is $x = \sqrt{3x}$, and square again.

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

Check $x = 0$: $\sqrt{1} - \sqrt{0} = 1 - 0 = 1$, which is correct. Check $x = 3$: $\sqrt{16} - \sqrt{9} = 4 - 3 = 1$, which is also correct. Both candidates survive, so both are solutions.