Solving Radical Equations: Practice

The radical is isolated, so square both sides and collect terms.

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

The candidates are $x = 3$ and $x = -2$.

Check $x = 3$: the left side is $\sqrt{9} = 3$ and the right side is $3$, so it survives.

Check $x = -2$: the left side is $\sqrt{4} = 2$ and the right side is $-2$, so $2 \ne -2$ and it is extraneous. The radicand $4$ is positive, so the failure is one of sign, not domain. Only $x = 3$ solves the equation.

Square both sides and collect terms.

$5x + 6 = x^2 + 4x + 4 \;\Longrightarrow\; x^2 - x - 2 = 0 \;\Longrightarrow\; (x - 2)(x + 1) = 0$

The candidates are $x = 2$ and $x = -1$. Both must be tested; do not assume one of them is an impostor.

Check $x = 2$: the left side is $\sqrt{16} = 4$ and the right side is $4$, so it survives.

Check $x = -1$: the left side is $\sqrt{1} = 1$ and the right side is $1$, so it survives too. Here the right side stayed non-negative at both candidates, so neither one is extraneous, and both are solutions.

Square both sides and collect terms.

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

The candidates are $x = 1$ and $x = 4$.

Check $x = 4$: the left side is $\sqrt{4} = 2$ and the right side is $2$, so it survives.

Check $x = 1$: the left side is $\sqrt{1} = 1$ and the right side is $-1$, so it is extraneous. Only $x = 4$ solves the equation.

The equation is a quadratic in $\sqrt{x}$, since $x = \left(\sqrt{x}\right)^2$. Let $u = \sqrt{x}$, keeping in mind that $u \ge 0$.

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

So $u = 3$ or $u = -2$. The value $u = -2$ is impossible, because a principal square root is never negative, so discard it at once.

From $\sqrt{x} = 3$ we get $x = 9$. Check in the original: $9 - \sqrt{9} - 6 = 9 - 3 - 6 = 0$, which is correct. So $x = 9$.

Isolate one radical, then square. The square of the binomial on the right keeps a cross term, which is where the surviving radical comes from.

$\sqrt{x + 3} = 1 + \sqrt{x - 2} \;\Longrightarrow\; x + 3 = 1 + 2\sqrt{x - 2} + (x - 2)$

The $x$ cancels from both sides, leaving $4 = 2\sqrt{x - 2}$, so $\sqrt{x - 2} = 2$ and $x - 2 = 4$, giving $x = 6$.

Check in the original: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$, which is correct. So $x = 6$.

The exponent $\tfrac{1}{2}$ is a square root, so this is $\sqrt{x + 4} = x - 8$. Raise both sides to the power $2$.

$x + 4 = x^2 - 16x + 64 \;\Longrightarrow\; x^2 - 17x + 60 = 0 \;\Longrightarrow\; (x - 5)(x - 12) = 0$

The candidates are $x = 5$ and $x = 12$.

Check $x = 12$: the left side is $(16)^{1/2} = 4$ and the right side is $4$, so it survives.

Check $x = 5$: the left side is $(9)^{1/2} = 3$ and the right side is $-3$, so it is extraneous. Only $x = 12$ solves the equation.

Split the exponent rather than guessing. Write $(x-1)^{3/2} = \left[(x-1)^{1/2}\right]^3$ and name the inner piece $u = (x - 1)^{1/2}$, so that $u \ge 0$.

$u^3 = 8 \;\Longrightarrow\; u = 2$

Cubing is reversible, so $u = 2$ is the only possibility. Then $(x - 1)^{1/2} = 2$ gives $x - 1 = 4$, so $x = 5$.

Check in the original: $(5 - 1)^{3/2} = 4^{3/2} = \left(\sqrt{4}\right)^3 = 2^3 = 8$, which is correct.

Square both sides and solve the resulting quadratic.

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

The candidates are $x = 4$ and $x = -1$.

Check $x = 4$: the radicand is $16 - 12 = 4$, so the left side is $\sqrt{4} = 2$, which matches.

Check $x = -1$: the radicand is $1 + 3 = 4$, so the left side is $\sqrt{4} = 2$, which also matches. The right side is the positive constant $2$, so there is no sign for squaring to spoil, and both candidates are genuine solutions.

Square both sides. The $x^2$ terms cancel, which turns a hard-looking equation into a linear one.

$x^2 + 9 = x^2 + 2x + 1 \;\Longrightarrow\; 8 = 2x \;\Longrightarrow\; x = 4$

Check in the original: the left side is $\sqrt{16 + 9} = \sqrt{25} = 5$ and the right side is $4 + 1 = 5$, which is correct. So $x = 4$ is the only solution.

Test each candidate in the original equation, one at a time.

At $x = 4$ the left side is $\sqrt{16} = 4$ and the right side is $4$, so this candidate is genuine.

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

$1 \ne -1$

So $x = -1$ is the extraneous one. It is not out of the domain (its radicand is $1$); it simply has the wrong sign on the right, and it is really a solution of the twin equation $\sqrt{3x + 4} = -x$.

Square both sides. On the left, remember that the coefficient $2$ is squared as well.

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

The only candidate is $x = 0$, a repeated root.

Check in the original: the left side is $2\sqrt{1} = 2$ and the right side is $0 + 2 = 2$, which is correct. So $x = 0$.

Square both sides. Writing $t = x + 1$ keeps the algebra tidy, since the right side is $-t$.

$t = t^2 \;\Longrightarrow\; t(t - 1) = 0 \;\Longrightarrow\; t = 0 \text{ or } t = 1$

So the candidates are $x = -1$ and $x = 0$.

Check $x = -1$: the left side is $\sqrt{0} = 0$ and the right side is $-(-1) - 1 = 0$, so it survives.

Check $x = 0$: the left side is $\sqrt{1} = 1$ and the right side is $-1$, so it is extraneous. Only $x = -1$ solves the equation, and it works because it makes both sides zero, the one place where a non-negative left side can meet a non-positive right side.