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The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

# Solved Examples - Radical Equations

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Specify the values that are excluded from the domain.
Write the domain in interval notation.
Solve, showing all work.
Check all solutions.
Identify any extraneous root(s).

(1.) $\sqrt{3x} = 9$

$\sqrt{3x} = 9$

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 $\underline{RHS} \\[3ex] 9$
(2.) ACT For what value of $x$ is the equation $\sqrt{x} + \sqrt{9} = \sqrt{36}$ true?

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

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 $\underline{RHS} \\[3ex] \sqrt{36} \\[3ex] 6$
(3.) $\sqrt{2p - 2} + \sqrt{p} = 7$

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

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. $\underline{RHS} \\[3ex] 7$
(4.) $\sqrt{2(x + 27)} - 3 = x$

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

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 $\underline{RHS} \\[3ex] x \\[3ex] -9$ $x \\[3ex] 5$
(5.) $\sqrt[3]{3x - 3} = -3$

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

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 $\underline{RHS} \\[3ex] -3$
(6.) $8x^{\dfrac{2}{3}} - 47x^{\dfrac{1}{3}} - 6 = 0$

$8x^{\dfrac{2}{3}} - 47x^{\dfrac{1}{3}} - 6 = 0 \\[5ex] is\:\: the\:\: same\:\: as \\[3ex] \left(\sqrt[3]{8x}\right)^2 - \sqrt[3]{47x} - 6 = 0 \\[5ex] also\:\: the\:\: same\:\: as \\[3ex] \sqrt[3]{(8x)^2} - \sqrt[3]{47x} - 6 = 0 \\[3ex]$ 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. $\underline{RHS} \\[3ex] 0$
(7.) $\sqrt{2x + 5} - \sqrt{x + 14} = -1$

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

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. $\underline{RHS} \\[3ex] -1$
(8.) $\sqrt{5 - 4\sqrt{x}} = \sqrt{x}$

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

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. $\underline{RHS} \\[3ex] \sqrt{x} \\[3ex] x = 1 \\[3ex] \sqrt{1} = 1$ $\sqrt{x} \\[3ex] x = 25 \\[3ex] \sqrt{25} = 5$
(9.) $\sqrt{2p + 6} - \sqrt{7 - 2p} = 1$

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

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. $\underline{RHS} \\[3ex] 1$
(10.) $\sqrt{p + 2} - 6 = 4$

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

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 $\underline{RHS} \\[3ex] 4$
(11.) ACT What is the solution set of

$\sqrt[4]{x^2 + 6x} = 2 \\[3ex] A.\:\: \{2\} \\[3ex] B.\:\: \{8\} \\[3ex] C.\:\: \{-8, 2\} \\[3ex] D.\:\: \{-2, 8\} \\[3ex] E.\:\: \{-3\pm\sqrt{13}\} \\[3ex]$

$\sqrt[4]{x^2 + 6x} = 2 \\[3ex] Raise\:\:both\:\:sides\:\:to\:\:exponent\:\:4 \\[3ex] (\sqrt[4]{x^2 + 6x})^4 = 2^4 \\[3ex] x^2 + 6x = 16 \\[3ex] x^2 + 6x - 16 = 0 \\[3ex] (x + 8)(x - 2) = 0 \\[3ex] x + 8 = 0 \:\:OR\:\: x - 2 = 0 \\[3ex] x = -8 \:\:OR\:\: x = 2 \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \sqrt[4]{x^2 + 6x} \\[3ex] x = -8 \\[3ex] = \sqrt[4]{(-8)^2 + 6(-8)} \\[3ex] = \sqrt[4]{64 - 48} \\[3ex] = \sqrt[4]{16} \\[3ex] = 2 \\[3ex] x = -8 \:\:is\:\:a\:\:solution$ $\underline{LHS} \\[3ex] x = 2 \\[3ex] = \sqrt[4]{(2)^2 + 6(2)} \\[3ex] = \sqrt[4]{4 + 12} \\[3ex] = \sqrt[4]{16} \\[3ex] = 2 \\[3ex] x = 2 \:\:is\:\:a\:\:solution$ $\underline{RHS} \\[3ex] 2$
(12.)

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

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