Solved Examples: Radical Equations

$\sqrt{3x} = 9$

The values excluded from the domain are all negative numbers.
All non-negative real numbers, $0$ and the positive numbers are included.

$ D = [0, \infty) \\[3ex] \sqrt{3x} = 9 \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sqrt{3x})^2 = 9^2 \\[3ex] 3x = 81 \\[3ex] x = \dfrac{81}{3} \\[5ex] x = 27 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt{3x} \\[3ex] x = 27 \\[3ex] = \sqrt{3 * 27} \\[3ex] = \sqrt{81} \\[3ex] = 9 \\[3ex] $ $\color{black}{x = 27}$ is a solution

$ \underline{RHS} \\[3ex] 9 $

$\sqrt{x} + \sqrt{9} = \sqrt{36}$

The values excluded from the domain are all negative numbers.
All non-negative real numbers, $0$ and the positive numbers are included.

$ D = [0, \infty) \\[3ex] \sqrt{x} + \sqrt{9} = \sqrt{36} \\[3ex] \sqrt{x} + 3 = 6 \\[3ex] \sqrt{x} = 6 - 3 \\[3ex] \sqrt{x} = 3 \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sqrt{x})^2 = 3^2 \\[3ex] x = 9 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt{x} + \sqrt{9} \\[3ex] x = 9 \\[3ex] \sqrt{9} + \sqrt{9} \\[3ex] = 3 + 3 \\[3ex] = 6 \\[3ex] $ $\color{black}{x = 9}$ is a solution

$ \underline{RHS} \\[3ex] \sqrt{36} \\[3ex] 6 $

$\sqrt{2p - 2} + \sqrt{p} = 7$

To find the domain;

$ 2p - 2 \ge 0 \\[3ex] 2p \ge 0 + 2 \\[3ex] 2p \ge 2 \\[3ex] p \ge 1 \\[3ex] Also: \\[3ex] p \ge 0 \\[3ex] Combine\:\: both\:\: (who\:\: satisfies\:\: both) \\[3ex] p \ge 1 \\[3ex] $ The values excluded from the domain are all real numbers less than $1$

$ D = [1, \infty) \\[3ex] \sqrt{2p - 2} + \sqrt{p} = 7 \\[3ex] \sqrt{2p - 2} = 7 - \sqrt{p} \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sqrt{2p - 2})^2 = (7 - sqrt{p})^2 \\[3ex] 2p - 2 = (7 - \sqrt{p})(7 - \sqrt{p}) \\[3ex] 2p - 2 = 49 - 7\sqrt{p} - 7\sqrt{p} + (\sqrt{p})^2 \\[3ex] 2p - 2 = 49 - 14\sqrt{p} + p \\[3ex] 14\sqrt{p} = 49 + p - 2p + 2 \\[3ex] 14\sqrt{p} = 51 - p \\[3ex] Square\:\: both\:\: sides\:\: again \\[3ex] (14\sqrt{p})^2 = (51 - p)^2 \\[3ex] 14^2 * (\sqrt{p})^2 = (51 - p)(51 - p) \\[3ex] 196p = 2601 - 51p - 51p + p^2 \\[3ex] 0 = 2601 - 102p + p^2 - 196p \\[3ex] 0 = p^2 -298p + 2601 \\[3ex] p^2 - 298p + 2601 = 0 \\[3ex] (p - 9)(p - 289) = 0 \\[3ex] p - 9 = 0 \:\:OR\:\: p - 289 = 0 \\[3ex] p = 9 \:\:OR\:\: p = 289 \\[3ex] $ Check
Check for both values.

$ \underline{LHS} \\[3ex] \sqrt{2p - 2} + \sqrt{p} \\[3ex] p = 9 \\[3ex] \sqrt{2(9) - 2} + \sqrt{9} \\[3ex] = \sqrt{18 - 2} + 3 \\[3ex] = \sqrt{16} + 3 \\[3ex] = 4 + 3 \\[3ex] = 7 \\[3ex] $ $\color{black}{p = 9}$ is a solution $ \sqrt{2p - 2} + \sqrt{p} \\[3ex] p = 289 \\[3ex] \sqrt{2(289) - 2} + \sqrt{289} \\[3ex] = \sqrt{578 - 2} + 17 \\[3ex] = \sqrt{576} + 17 \\[3ex] = 24 + 17 \\[3ex] = 41 \\[3ex] $ $p = 289$ is NOT a solution. $p = 289$ is an extraneous solution.

$ \underline{RHS} \\[3ex] 7 $

$\sqrt{2(x + 27)} - 3 = x$

To find the domain;

$ 2(x + 27) \ge 0 \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: 2 \\[3ex] x + 27 \ge 0 \\[3ex] x \ge 0 - 27 \\[3ex] x \ge -27 \\[3ex] $ The values excluded from the domain are all real numbers less than $-27$

$ D = [-27, \infty) \\[3ex] \sqrt{2(x + 27)} - 3 = x \\[3ex] \sqrt{2(x + 27)} = x + 3 \\[3ex] Square\:\: both\:\: sides \\[3ex] \left(\sqrt{2(x + 27)}\right)^2 = (x + 3)^2 \\[3ex] 2(x + 27) = (x + 3)(x + 3) \\[3ex] 2x + 54 = x^2 + 3x + 3x + 9 \\[3ex] 2x + 54 = x^2 + 6x + 9 \\[3ex] 0 = x^2 + 6x + 9 - 2x - 54 \\[3ex] 0 = x^2 + 4x - 45 \\[3ex] x^2 + 4x - 45 = 0 \\[3ex] (x + 9)(x - 5) = 0 \\[3ex] x + 9 = 0 \:\:OR\:\: x - 5 = 0 \\[3ex] x = -9 \:\:OR\:\: x = 5 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt{2(x + 27)} - 3 \\[3ex] x = -9 \\[3ex] \sqrt{2(x + 27)} - 3 \\[3ex] = \sqrt{2(-9 + 27)} - 3 \\[3ex] = \sqrt{2(18)} - 3 \\[3ex] = \sqrt{36} - 3 \\[3ex] = 6 - 3 \\[3ex] = 3 \\[3ex] $ $x = -9$ is NOT a solution $x = -9$ is an extraneous solution $ \sqrt{2(x + 27)} - 3 \\[3ex] x = 5 \\[3ex] \sqrt{2(5 + 27)} - 3 \\[3ex] = \sqrt{2(32)} - 3 \\[3ex] = \sqrt{64} - 3 \\[3ex] = 8 - 3 \\[3ex] = 5 \\[3ex] $ $\color{black}{x = 5}$ is a solution

$ \underline{RHS} \\[3ex] x \\[3ex] -9 $ $ x \\[3ex] 5 $

$\sqrt[3]{3x - 3} = -3$

There are no values excluded from the domain because of the cube root.
All real numbers are included in the domain.
Any odd-th root of any number (positive, negative, or zero) has a result.

$ D = (-\infty, \infty) \\[3ex] \sqrt[3]{3x - 3} = -3 \\[3ex] Cube\:\: both\:\: sides \\[3ex] (\sqrt[3]{3x - 3})^3 = (-3)^3 \\[3ex] 3x - 3 = -27 \\[3ex] 3x = -27 + 3 \\[3ex] 3x = -24 \\[3ex] x = -\dfrac{24}{3} \\[5ex] x = -8 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt[3]{3x - 3} \\[3ex] x = -8 \\[3ex] \sqrt[3]{3(-8) - 3} \\[3ex] = \sqrt[3]{-24 - 3} \\[3ex] = \sqrt[3]{-27} \\[3ex] = -3 \\[3ex] $ $\color{black}{x = -8}$ is a solution

$ \underline{RHS} \\[3ex] -3 $

$ 8x^{\dfrac{2}{3}} - 47x^{\dfrac{1}{3}} - 6 = 0 \\[5ex] is\:\: the\:\: same\:\: as \\[3ex] \left(\sqrt[3]{8x}\right)^2 - \sqrt[3]{47x} - 6 = 0 \\[5ex] also\:\: the\:\: same\:\: as \\[3ex] \sqrt[3]{(8x)^2} - \sqrt[3]{47x} - 6 = 0 \\[3ex] $ There are no values excluded from the domain because of the cube root.
All real numbers are included in the domain.
Any odd-th root of any number (positive, negative, or zero) has a result.

$D = (-\infty, \infty)$

As we see, cube root is common.
Looking at the original equation "as is", it is difficult to solve.
So, let us re-write the equation in a way we can solve it.
We need to do some substitution
Anytime you see an equation like this, it is better to find some sort of "substitution".

$ 8x^{\dfrac{2}{3}} - 47x^{\dfrac{1}{3}} - 6 = 0 \\[5ex] 8 * x^{\dfrac{1}{3} * 2} - 47 * x^{\dfrac{1}{3}} - 6 = 0 \\[5ex] 8 * \left(x^{\dfrac{1}{3}}\right)^2 - 47 * x^{\dfrac{1}{3}} - 6 = 0 \\[5ex] Let\:\: x^{\dfrac{1}{3}} = p \\[5ex] 8 * p^2 - 47 * p - 6 = 0 \\[3ex] 8p^2 - 47p - 6 = 0 \\[3ex] (8p^2)(-6) = -48p^2 \\[3ex] Factors\:\: are\:\: p \:\:and\:\: -48p \\[3ex] 8p^2 - 47p - 6 = 0 \\[3ex] = 8p^2 + p - 48p - 6 = 0 \\[3ex] = p(8p + 1) - 6(8p + 1) = 0 \\[3ex] (8p + 1) = 0 \:\:OR\:\: (p - 6) = 0 \\[3ex] 8p + 1 = 0 \\[3ex] 8p = -1 \\[3ex] p = -\dfrac{1}{8} \\[5ex] p - 6 = 0 \\[3ex] p = 6 \\[3ex] p = -\dfrac{1}{8} \:\:OR\:\: p = 6 \\[5ex] But\:\: x^{\dfrac{1}{3}} = p \\[5ex] When\:\: p = -\dfrac{1}{8} \\[5ex] x^{\dfrac{1}{3}} = -\dfrac{1}{8} \\[5ex] Cube\:\: both\:\: sides \\[3ex] \left(x^{\dfrac{1}{3}}\right)^3 = \left(-\dfrac{1}{8}\right)^3 \\[5ex] x = \dfrac{(-1)^3}{8^3} \\[5ex] x = -\dfrac{1}{512} \\[5ex] When\:\: p = 6 \\[3ex] x^{\dfrac{1}{3}} = 6 \\[5ex] Cube\:\: both\:\: sides \\[3ex] \left(x^{\dfrac{1}{3}}\right)^3 = 6^3 \\[5ex] x = 216 \\[3ex] x = -\dfrac{1}{512} \:\:OR\:\: x = 216 \\[5ex] $ Check

$ \underline{LHS} \\[3ex] 8x^{\dfrac{2}{3}} - 47x^{\dfrac{1}{3}} - 6 \\[5ex] x = -\dfrac{1}{512} \\[5ex] 8 * x^{\dfrac{2}{3}} - 47 * x^{\dfrac{1}{3}} - 6 \\[5ex] = 8 * \left(-\dfrac{1}{512}\right)^{\dfrac{2}{3}} - 47 * \left(-\dfrac{1}{512}\right)^{\dfrac{1}{3}} - 6 \\[5ex] = 8 * \left(\sqrt[3]{-\dfrac{1}{512}}\right)^2 - 47 * \sqrt[3]{-\dfrac{1}{512}} - 6 \\[5ex] = 8 * \left(-\dfrac{1}{8}\right)^2 - 47 * -\dfrac{1}{8} - 6 \\[5ex] = 8 * \dfrac{1}{64} + \dfrac{47}{8} - 6 \\[5ex] = \dfrac{1}{8} + \dfrac{47}{8} - \dfrac{48}{8} \\[5ex] = \dfrac{1 + 47 - 48}{8} \\[5ex] = \dfrac{0}{8} \\[5ex] = 0 \\[3ex] $ $\color{black}{x = -\dfrac{1}{512}}$ is a solution $ 8x^{\dfrac{2}{3}} - 47x^{\dfrac{1}{3}} - 6 \\[5ex] x = 216 \\[3ex] 8 * 216^{\dfrac{2}{3}} - 47 * 216^{\dfrac{1}{3}} - 6 \\[5ex] = 8 * \sqrt[3]{216}^2 - 47 * \sqrt[3]{216} - 6 \\[3ex] = 8 * 6^2 - 47 * 6 - 6 \\[3ex] = 8 * 36 - 282 - 6 \\[3ex] = 288 - 282 - 6 \\[3ex] = 0 \\[3ex] $ $\color{black}{x = 216}$ is a solution.

$ \underline{RHS} \\[3ex] 0 $

$\sqrt{2x + 5} - \sqrt{x + 14} = -1$

To find the domain;

$ 2x + 5 \ge 0 \\[3ex] 2x \ge 0 - 5 \\[3ex] 2x \ge -5 \\[3ex] x \ge -\dfrac{5}{2} \\[5ex] Also: \\[3ex] x + 14 \ge 0 \\[3ex] x \ge -14 \\[3ex] Combine\:\: both\:\: (who\:\: satisfies\:\: both) \\[3ex] x \ge -\dfrac{5}{2} \\[5ex] $ The values excluded from the domain are all real numbers less than $-\dfrac{5}{2}$

$ D = \left[-\dfrac{5}{2}, \infty\right) \\[5ex] \sqrt{2x + 5} - \sqrt{x + 14} = -1 \\[3ex] \sqrt{2x + 5} = -1 + \sqrt{x + 14} \\[3ex] Square\:\: both\:\: sides \\[3ex] \left(\sqrt{2x + 5}\right)^2 = \left(-1 + \sqrt{x + 14}\right)^2 \\[3ex] 2x + 5 = (-1 + \sqrt{x + 14})(-1 + \sqrt{x + 14}) \\[3ex] (-1 + \sqrt{x + 14})(-1 + \sqrt{x + 14}) \\[3ex] (-1)(-1) = 1 \\[3ex] (-1)(\sqrt{x + 14}) = -\sqrt{x + 14} \\[3ex] \sqrt{x + 14}(-1) = -\sqrt{x + 14} \\[3ex] (\sqrt{x + 14})(\sqrt{x + 14}) = \left(\sqrt{x + 14}\right)^2 = x + 14 \\[3ex] \implies 2x + 5 = 1 - \sqrt{x + 14} - \sqrt{x + 14} + (x + 14) \\[3ex] 2x + 5 = 1 - 2\sqrt{x + 14} + x + 14 \\[3ex] 2\sqrt{x + 14} = 1 + x + 14 - 2x - 5 \\[3ex] 2\sqrt{x + 14} = 10 - x \\[3ex] Square\:\: both\:\: sides\:\: again \\[3ex] \left(2\sqrt{x + 14}\right)^2 = (10 - x)^2 \\[3ex] 2^2 * (\sqrt{x + 14})^2 = (10 - x)(10 - x) \\[3ex] 4(x + 14) = 100 - 10x - 10x + x^2 \\[3ex] 4x + 56 = 100 - 20x + x^2 \\[3ex] 100 - 20x + x^2 = 4x + 56 \\[3ex] x^2 - 20x + 100 - 4x - 56 = 0 \\[3ex] x^2 - 24x + 44 = 0 \\[3ex] (x - 2)(x - 22) = 0 \\[3ex] x - 2 = 0 \:\:OR\:\: x - 22 = 0 \\[3ex] x = 2 \:\:OR\:\: x = 22 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt{2x + 5} - \sqrt{x + 14} \\[3ex] x = 2 \\[3ex] \sqrt{2x + 5} - \sqrt{x + 14} \\[3ex] \sqrt{2(2) + 5} - \sqrt{2 + 14} \\[3ex] = \sqrt{4 + 5} - \sqrt{16} \\[3ex] = \sqrt{9} - \sqrt{16} \\[3ex] = 3 - 4 \\[3ex] = -1 \\[3ex] $ $\color{black}{x = 2}$ is a solution $ \sqrt{2x + 5} - \sqrt{x + 14} \\[3ex] x = 22 \\[3ex] \sqrt{2x + 5} - \sqrt{x + 14} \\[3ex] \sqrt{2(22) + 5} - \sqrt{22 + 14} \\[3ex] = \sqrt{44 + 5} - \sqrt{36} \\[3ex] = \sqrt{49} - \sqrt{36} \\[3ex] = 7 - 6 \\[3ex] = 1 \\[3ex] $ $x = 22$ is NOT a solution. $x = 22$ is an extraneous solution.

$ \underline{RHS} \\[3ex] -1 $

$\sqrt{5 - 4\sqrt{x}} = \sqrt{x}$

To find the domain;

$ x \ge 0 \\[3ex] $ The values excluded from the domain are all real numbers less than $0$

$ D = [0, \infty) \\[3ex] \sqrt{5 - 4\sqrt{x}} = \sqrt{x} \\[3ex] Square\:\: both\:\: sides \\[3ex] \left(\sqrt{5 - 4\sqrt{x}}\right)^2 = (\sqrt{x})^2 \\[3ex] 5 - 4\sqrt{x} = x \\[3ex] 5 - x = 4\sqrt{x} \\[3ex] 4\sqrt{x} = 5 - x \\[3ex] Square\:\: both\:\: sides\:\: again \\[3ex] (4\sqrt{x})^2 = (5 - x)^2 \\[3ex] 4^2 * (\sqrt{x})^2 = (5 - x)(5 - x) \\[3ex] 16 * x = 25 - 5x - 5x + x^2 \\[3ex] 16x = 25 - 10x + x^2 \\[3ex] 0 = x^2 - 10x - 16x + 25 \\[3ex] 0 = x^2 - 26x + 25 \\[3ex] x^2 - 26x + 25 = 0 \\[3ex] (x - 1)(x - 25) = 0 \\[3ex] x - 1 = 0 \:\:OR\:\: x - 25 = 0 \\[3ex] x = 1 \:\:OR\:\: x = 25 \\[3ex] $ Check
Check for both values.

$ \underline{LHS} \\[3ex] \sqrt{5 - 4\sqrt{x}} \\[3ex] x = 1 \\[3ex] \sqrt{5 - 4\sqrt{1}} \\[3ex] = \sqrt{5 - 4(1)} \\[3ex] = \sqrt{5 - 4} \\[3ex] = \sqrt{1} \\[3ex] = 1 \\[3ex] $ $\color{black}{x = 1}$ is a solution $ \sqrt{5 - 4\sqrt{x}} \\[3ex] x = 25 \\[3ex] \sqrt{5 - 4\sqrt{25}} \\[3ex] = \sqrt{5 - 4(5)} \\[3ex] = \sqrt{5 - 20} \\[3ex] = \sqrt{-15} \\[3ex] = NAN(Not\:\: a\:\: real\:\: number) \\[3ex] $ $x = 25$ is NOT a solution. $x = 25$ is an extraneous solution.

$ \underline{RHS} \\[3ex] \sqrt{x} \\[3ex] x = 1 \\[3ex] \sqrt{1} = 1 $ $ \sqrt{x} \\[3ex] x = 25 \\[3ex] \sqrt{25} = 5 $

$\sqrt{2p + 6} - \sqrt{7 - 2p} = 1$

To find the domain;

$ 2p + 6 \ge 0 \\[3ex] 2p \ge 0 - 6 \\[3ex] 2p \ge -6 \\[3ex] p \ge -\dfrac{6}{2} \\[5ex] p \ge -3 \\[3ex] Also: \\[3ex] 7 - 2p \ge 0 \\[3ex] 7 \ge 2p \\[3ex] 2p \le 7 \\[3ex] p \le \dfrac{7}{2} \\[5ex] Combine\:\: both\:\: (who\:\: satisfies\:\: both) \\[3ex] p \ge -3 \\[3ex] -3 \le p \\[3ex] p \le \dfrac{7}{2} \\[5ex] -3 \le p \le \dfrac{7}{2} \\[5ex] $ The values excluded from the domain are all real numbers less than $-3$ and all real numbers greater than $\dfrac{7}{2}$

$ D = \left[-3, \dfrac{7}{2}\right] \\[5ex] \sqrt{2p + 6} - \sqrt{7 - 2p} = 1 \\[3ex] \sqrt{2p + 6} = 1 + \sqrt{7 - 2p} \\[3ex] Square\:\: both\:\: sides \\[3ex] \left(\sqrt{2p + 6}\right)^2 = \left(1 + \sqrt{7 - 2p}\right)^2 \\[3ex] 2p + 6 = (1 + \sqrt{7 - 2p})(1 + \sqrt{7 - 2p}) \\[3ex] (1 + \sqrt{7 - 2p})(1 + \sqrt{7 - 2p}) \\[3ex] (1)(1) = 1 \\[3ex] (1)(\sqrt{7 - 2p}) = \sqrt{7 - 2p} \\[3ex] \sqrt{7 - 2p}(1) = \sqrt{7 - 2p} \\[3ex] (\sqrt{7 - 2p})(\sqrt{7 - 2p}) = \left(\sqrt{7 - 2p}\right)^2 = 7 - 2p \\[3ex] \implies 2p + 6 = 1 + \sqrt{7 - 2p} + \sqrt{7 - 2p} + (7 - 2p) \\[3ex] 2p + 6 = 1 + 2\sqrt{7 - 2p} + 7 - 2p \\[3ex] 2p + 6 = 2\sqrt{7 - 2p} - 2p + 8 \\[3ex] 2\sqrt{7 - 2p} - 2p + 8 = 2p + 6 \\[3ex] 2\sqrt{7 - 2p} = 2p + 6 + 2p - 8 \\[3ex] 2\sqrt{7 - 2p} = 4p - 2 \\[3ex] 2\sqrt{7 - 2p} = 2(2p - 1) \\[3ex] Divide \:\:both\:\: sides\:\: 2 \\[3ex] \dfrac{2\sqrt{7 - 2p}}{2} = \dfrac{2(2p - 1)}{2} \\[5ex] \sqrt{7 - 2p} = 2p - 1 \\[3ex] Square\:\: both\:\: sides\:\: again \\[3ex] \left(\sqrt{7 - 2p}\right)^2 = (2p - 1)^2 \\[3ex] 7 - 2p = (2p - 1)(2p - 1) \\[3ex] 7 - 2p = 4p^2 - 2p - 2p + 1 \\[3ex] 7 - 2p = 4p^2 - 4p + 1 \\[3ex] 0 = 4p^2 - 4p + 1 - 7 + 2p \\[3ex] 0 = 4p^2 - 2p - 6 \\[3ex] 4p^2 - 2p - 6 = 0 \\[3ex] 2(2p^2 - p - 3) = 0 \\[3ex] Divide \:\:both\:\: sides\:\: 2 \\[3ex] \dfrac{2(2p^2 - p - 3)}{2} = \dfrac{0}{2} \\[3ex] 2p^2 - p - 3 = 0 \\[3ex] 2p^2 + 2p - 3p - 3 = 0 \\[3ex] 2p(p + 1) - 3(p + 1) = 0 \\[3ex] p + 1 = 0 \:\:OR\:\: 2p - 3 = 0 \\[3ex] p = -1 \:\:OR\:\: 2p = 3 \\[3ex] p = -1 \:\:OR\:\: p = \dfrac{3}{2} \\[5ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt{2p + 6} - \sqrt{7 - 2p} \\[3ex] p = -1 \\[5ex] \sqrt{2p + 6} - \sqrt{7 - 2p} \\[3ex] \sqrt{2(-1) + 6} - \sqrt{7 - 2(-1)} \\[3ex] = \sqrt{-2 + 6} - \sqrt{7 + 2} \\[3ex] = \sqrt{4} - \sqrt{9} \\[3ex] = 2 - 3 \\[3ex] = -1 \\[3ex] $ $p = -1$ is NOT a solution. $p = -1$ is an extraneous solution. $ \sqrt{2p + 6} - \sqrt{7 - 2p} \\[3ex] p = \dfrac{3}{2} \\[5ex] \sqrt{2\left(\dfrac{3}{2}\right) + 6} - \sqrt{7 - 2\left(\dfrac{3}{2}\right)} \\[5ex] \sqrt{3 + 6} - \sqrt{7 - 3} \\[3ex] = \sqrt{9} - \sqrt{4} \\[3ex] = 3 - 2 \\[3ex] = 1 \\[3ex] $ $\color{black}{p = \dfrac{3}{2}}$ is a solution.

$ \underline{RHS} \\[3ex] 1 $

$\sqrt{p + 2} - 6 = 4$

To find the domain;

$ p + 2 \ge 0 \\[3ex] p \ge -2 \\[3ex] $ The values excluded from the domain are all real numbers less than $-2$

$ D = [-2, \infty) \\[3ex] \sqrt{p + 2} - 6 = 4 \\[3ex] \sqrt{p + 2} = 4 + 6 \\[3ex] \sqrt{p + 2} = 10 \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sqrt{p + 2})^2 = 10^2 \\[3ex] p + 2 = 100 \\[3ex] p = 100 - 2 \\[3ex] p = 98 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt{p + 2} - 6 \\[3ex] p = 98 \\[3ex] \sqrt{98 + 2} - 6 \\[3ex] = \sqrt{100} - 6 \\[3ex] = 10 - 6 \\[3ex] = 4 \\[3ex] $ $\color{black}{p = 98}$ is a solution

$ \underline{RHS} \\[3ex] 4 $

$ \sqrt[4]{x^2 + 6x} = 2 \\[3ex] Raise\:\:both\:\:sides\:\:to\:\:exponent\:\:4 \\[3ex] (\sqrt[4]{x^2 + 6x})^4 = 2^4 \\[3ex] x^2 + 6x = 16 \\[3ex] x^2 + 6x - 16 = 0 \\[3ex] (x + 8)(x - 2) = 0 \\[3ex] x + 8 = 0 \:\:OR\:\: x - 2 = 0 \\[3ex] x = -8 \:\:OR\:\: x = 2 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] \sqrt[4]{x^2 + 6x} \\[3ex] x = -8 \\[3ex] = \sqrt[4]{(-8)^2 + 6(-8)} \\[3ex] = \sqrt[4]{64 - 48} \\[3ex] = \sqrt[4]{16} \\[3ex] = 2 \\[3ex] x = -8 \:\:is\:\:a\:\:solution $ $ \underline{LHS} \\[3ex] x = 2 \\[3ex] = \sqrt[4]{(2)^2 + 6(2)} \\[3ex] = \sqrt[4]{4 + 12} \\[3ex] = \sqrt[4]{16} \\[3ex] = 2 \\[3ex] x = 2 \:\:is\:\:a\:\:solution $

$ \underline{RHS} \\[3ex] 2 $

$ \sqrt{(3 + \sqrt{x})} = 1 + \sqrt{2} \\[3ex] Square\;\;both\;\;sides \\[3ex] (\sqrt{(3 + \sqrt{x})})^2 = (1 + \sqrt{2})^2 \\[3ex] 3 + \sqrt{x} = (1 + \sqrt{2})(1 + \sqrt{2}) \\[3ex] 3 + \sqrt{x} = 1 + \sqrt{2} + \sqrt{2} + 2 \\[3ex] 3 + \sqrt{x} = 3 + 2\sqrt{2} \\[3ex] \sqrt{x} = 3 + 2\sqrt{2} - 3 \\[3ex] \sqrt{x} = 2\sqrt{2} \\[3ex] Square\;\;both\;\;sides \\[3ex] (\sqrt{x})^2 = (2\sqrt{2})^2 \\[3ex] x = 2^2 * (\sqrt{2})^2 \\[3ex] x = 4 * 2 \\[3ex] x = 8 $