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It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

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

# Solved Examples: Radical Equations

As applicable, verify your answers with the Expressions and Equations Calculators

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

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

As applicable (unless specifie otherwise),
For each radical equation:
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 $\sqrt[4]{x^2 + 6x} = 2$

$A.\:\: \{2\} \\[3ex] B.\:\: \{8\} \\[3ex] C.\:\: \{-8, 2\} \\[3ex] D.\:\: \{-2, 8\} \\[3ex] E.\:\: \{-3\pm\sqrt{13}\} \\[3ex]$

$\sqrt[4]{x^2 + 6x} = 2 \\[3ex] Raise\:\:both\:\:sides\:\:to\:\:exponent\:\:4 \\[3ex] (\sqrt[4]{x^2 + 6x})^4 = 2^4 \\[3ex] x^2 + 6x = 16 \\[3ex] x^2 + 6x - 16 = 0 \\[3ex] (x + 8)(x - 2) = 0 \\[3ex] x + 8 = 0 \:\:OR\:\: x - 2 = 0 \\[3ex] x = -8 \:\:OR\:\: x = 2 \\[3ex]$ 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.) ACT If $\sqrt{(3 + \sqrt{3})} = 1 + \sqrt{2}$, then x = ?

$F.\;\; 0 \\[3ex] G.\;\; 2 \\[3ex] H.\;\; 4 \\[3ex] J.\;\; 8 \\[3ex] K.\;\; 10 \\[3ex]$

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