Solving Radical Equations: Additional Practice (Set 2)

Isolate one radical and square. The binomial on the right leaves a cross term, so one radical survives.

$\sqrt{2x + 5} = 1 + \sqrt{x + 6} \;\Longrightarrow\; 2x + 5 = 1 + 2\sqrt{x + 6} + (x + 6)$

Simplify to $x - 2 = 2\sqrt{x + 6}$ and square again: $x^2 - 4x + 4 = 4x + 24$, so $x^2 - 8x - 20 = 0$ and $(x - 10)(x + 2) = 0$.

Check $x = 10$: $\sqrt{25} - \sqrt{16} = 5 - 4 = 1$, which is correct.

Check $x = -2$: $\sqrt{1} - \sqrt{4} = 1 - 2 = -1 \ne 1$, so it is extraneous. Both radicands were non-negative at $x = -2$, so it was inside the domain and failed purely on sign. Only $x = 10$ solves the equation.

Square both sides and collect terms.

$x + 1 = x^2 - 10x + 25 \;\Longrightarrow\; x^2 - 11x + 24 = 0 \;\Longrightarrow\; (x - 3)(x - 8) = 0$

Check $x = 8$: the left side is $\sqrt{9} = 3$ and the right side is $8 - 5 = 3$, so it survives.

Check $x = 3$: the left side is $\sqrt{4} = 2$ and the right side is $3 - 5 = -2$, so it is extraneous. Its radicand is $4$, so it is inside the domain; it fails on sign alone. Only $x = 8$ solves the equation.

Isolate one radical and square.

$\sqrt{3x + 1} = 2 + \sqrt{x - 1} \;\Longrightarrow\; 3x + 1 = 4 + 4\sqrt{x - 1} + (x - 1)$

Simplify to $2x - 2 = 4\sqrt{x - 1}$, then divide by $2$ to get $x - 1 = 2\sqrt{x - 1}$. Squaring again gives $(x - 1)^2 = 4(x - 1)$, so $(x - 1)(x - 5) = 0$.

Check $x = 1$: $\sqrt{4} - \sqrt{0} = 2 - 0 = 2$, which is correct. Check $x = 5$: $\sqrt{16} - \sqrt{4} = 4 - 2 = 2$, which is also correct. Neither candidate is extraneous, so both are solutions.

Cube both sides, which is a reversible step, and expand the cube on the right.

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

Factoring gives $(x - 2)(x + 1) = 0$, so the candidates are $x = 2$ and $x = -1$.

Check $x = 2$: the left side is $\sqrt[3]{1} = 1$ and the right side is $1$. Check $x = -1$: the left side is $\sqrt[3]{-8} = -2$ and the right side is $-2$. Both survive, and by the reversibility of cubing neither could have been extraneous.

Notice that $x^{1/2} = \left(x^{1/4}\right)^2$, so the equation is a quadratic in $x^{1/4}$. Let $u = x^{1/4}$, and remember that $u \ge 0$ because it is a principal fourth root.

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

So $u = 2$ or $u = -3$, and $u = -3$ is impossible. From $x^{1/4} = 2$ we get $x = 2^4 = 16$.

Check in the original: $16^{1/2} + 16^{1/4} = 4 + 2 = 6$, which is correct. So $x = 16$.

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

$u^2 - 5u + 6 = 0 \;\Longrightarrow\; (u - 2)(u - 3) = 0$

Both $u = 2$ and $u = 3$ are non-negative, so neither is rejected. Squaring gives $x = 4$ and $x = 9$.

Check $x = 4$: $4 - 5(2) + 6 = 0$. Check $x = 9$: $9 - 5(3) + 6 = 0$. Both are solutions. Note that $u = 2$ and $u = 3$ are values of $\sqrt{x}$, not of $x$, which is why the answers are $4$ and $9$ rather than $2$ and $3$.

Isolate the radical in each equation and read the sign of the right side.

In $\sqrt{x} + 5 = 2$, isolating gives $\sqrt{x} = 2 - 5 = -3$, and a principal square root is never negative, so no real $x$ works.

$\sqrt{x} = -3 \;\Longrightarrow\; \text{no solution}$

The other three are all solvable: $\sqrt{x} = 3$ gives $x = 9$; $\sqrt[3]{x} = -2$ gives $x = -8$, because a cube root is allowed to be negative; and $\sqrt{x - 1} = 0$ gives $x = 1$, because zero is non-negative.

Square both sides, expanding the binomial on the right.

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

One radical remains, so square again.

$x^2 + 4x + 4 = 4x \;\Longrightarrow\; x^2 + 4 = 0$

No real number has a square of $-4$, so there is no candidate at all, and the equation has no real solution. You can see it directly too: $\sqrt{2x + 3}$ stays above $\sqrt{x} + 1$ for every $x \ge 0$.

Squaring turns $A = B$ into the pair of statements $A = B$ or $A = -B$, because $A^2 = B^2$ factors as $(A - B)(A + B) = 0$. So a candidate can satisfy the squared equation by belonging to the twin.

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

Here the radicand is $9$, so $x = 2$ is safely inside the domain, yet the two sides have opposite signs and it is extraneous. A domain check would wrongly accept it. Substituting into the original tests exactly what a solution is required to do, so it catches every failure regardless of cause. (Raising both sides to a positive integer power never loses a solution, so the suggestion that a power can lose one describes a problem that cannot arise.)

Isolate one radical and square.

$\sqrt{x + 4} = 1 + \sqrt{x - 3} \;\Longrightarrow\; x + 4 = 1 + 2\sqrt{x - 3} + (x - 3)$

The $x$ cancels, leaving $6 = 2\sqrt{x - 3}$, so $\sqrt{x - 3} = 3$ and $x - 3 = 9$, giving $x = 12$.

Check in the original: $\sqrt{16} - \sqrt{9} = 4 - 3 = 1$, which is correct. So $x = 12$.

Square both sides and collect terms.

$3 - x = x^2 + 6x + 9 \;\Longrightarrow\; x^2 + 7x + 6 = 0 \;\Longrightarrow\; (x + 1)(x + 6) = 0$

Check $x = -1$: the left side is $\sqrt{4} = 2$ and the right side is $-1 + 3 = 2$, so it survives.

Check $x = -6$: the left side is $\sqrt{9} = 3$ and the right side is $-6 + 3 = -3$, so it is extraneous. A negative value of $x$ is not automatically wrong, as $x = -1$ shows; what matters is whether both sides agree.

Substitute $x = 3$ to pin down the constant.

$\sqrt{3 + a} = 3 \;\Longrightarrow\; 3 + a = 9 \;\Longrightarrow\; a = 6$

Now solve $\sqrt{x + 6} = x$ completely. Squaring gives $x + 6 = x^2$, so $x^2 - x - 6 = 0$ and $(x - 3)(x + 2) = 0$, with candidates $x = 3$ and $x = -2$.

Check $x = -2$: the left side is $\sqrt{4} = 2$ and the right side is $-2$, so it is extraneous. Only $x = 3$ survives, so the solution set contains $x = 3$ alone.