1. We told you back on page ?? that roots do not ‘distribute’ across addition and since \(x^{2} + 1\) cannot be factored over the real numbers, \(\sqrt {x^{2} + 1}\) cannot be simplified. It may seem silly to start with this example but it is extremely important that you understand what maneuvers are legal and which ones are not. ^[3] (You really do need to understand this otherwise horrible evil will plague your future studies in Math. If you say something totally wrong like \(\sqrt {x^{2} + 1} = x + 1\) then you may never pass Calculus. PLEASE be careful!)

2. Again we note that \(\sqrt {t^2-10t+25} \neq \sqrt {t^2} - \sqrt {10t} + \sqrt {25}\), since radicals do not distribute across addition and subtraction. ^[4] (Let \(t = 1\) and see what happens to \(\sqrt {t^2-10t+25}\) versus \(\sqrt {t^2} - \sqrt {10t} + \sqrt {25}\).) In this case, however, we can factor the radicand and simplify as

   \[ \sqrt {t^2 - 10t + 25} = \sqrt {(t-5)^2} = |t-5| \]
   Without knowing more about the value of \(t\), we have no idea if \(t-5\) is positive or negative so \(|t-5|\) is our final answer. ^[5] (In general, \(|t-5| \neq |t| - |5|\) and \(|t-5| \neq t + 5\) so watch what you’re doing!)

3. To simplify \(\sqrt [3]{48x^{14}}\), we need to look for perfect cubes in the radicand. For the cofficient, we have \(48 = 8 \cdot 6 = 2^3 \cdot 6\). To find the largest perfect cube factor in \(x^{14}\), we divide \(14\) (the exponent on \(x\)) by \(3\) (since we are looking for a perfect cube). We get \(4\) with a remainder of \(2\). This means \(14 = 4 \cdot 3 + 2\), so \(x^{14} = x^{4 \cdot 3 + 2} = x^{4 \cdot 3} x^2 = (x^4)^3 x^2\). Putting this altogether gives:

   \[\begin{array}{rclr} \sqrt [3]{48x^{14}} & = & \sqrt [3]{2^3 \cdot 6 \cdot (x^4)^3 x^2} & \text {Factor out perfect cubes} \\ & = & \sqrt [3]{2^3} \sqrt [3]{(x^4)^3} \sqrt [3]{6x^2} & \text {Rearrange factors, Product Rule of Radicals} \\ & = & 2x^4\sqrt [3]{6x^2} & \\ \end{array} \]

4. In this example, we are looking for perfect fourth powers in the radicand. In the numerator \(r^4\) is clearly a perfect fourth power. For the denominator, we take the power on the \(L\), namely \(12\), and divide by \(4\) to get \(3\). This means \(L^{8} = L^{2\cdot 4} = (L^2)^{4}\). We get

   \[ \begin{array}{rclr} \sqrt [4]{\frac {\pi r^{4}}{L^{12}}} & = & \frac {\sqrt [4]{\pi r^{4}}}{\sqrt [4]{L^{12}}} & \text {Quotient Rule of Radicals} \\ & = & \frac {\sqrt [4]{\pi }\sqrt [4]{r^{4}}}{\sqrt [4]{(L^2)^{4}}} & \text {Product Rule of Radicals} \\ & = & \frac {\sqrt [4]{\pi }|r|}{|L^2|} & \text {Simplify} \\ \end{array}\]
   Without more information about \(r\), we cannot simplify \(|r|\) any further. However, we can simplify \(|L^2|\). Regardless of the choice of \(L\), \(L^2 \geq 0\). Actually, \(L^2 > 0\) because \(L\) is in the denominator which means \(L \neq 0\). Hence, \(|L^2| = L^2\). Our answer simplifies to:
   \[ \frac {\sqrt [4]{\pi }|r|}{|L^2|} = \frac {|r|\sqrt [4]{\pi }}{L^2} \]

5. After a quick cancellation (two of the \(2\)’s in the second term) we need to obtain a common denominator. Since we can view the first term as having a denominator of \(1\), the common denominator is precisely the denominator of the second term, namely \((\sqrt [3]{x^2-4})^2\). With common denominators, we proceed to add the two fractions. Our last step is to factor the numerator to see if there are any cancellation opportunities with the denominator.

   \[ \begin{array}{rclr} 2x \sqrt [3]{x^2-4} + 2\left (\frac {1}{2(\sqrt [3]{x^2-4})^2}\right ) (2x) & = & 2x \sqrt [3]{x^2-4} + \cancel {2}\left (\frac {1}{\cancel {2}(\sqrt [3]{x^2-4})^2}\right ) (2x) & \text {Reduce}\\ & = & 2x \sqrt [3]{x^2-4} + \frac {2x}{(\sqrt [3]{x^2-4})^2} & \text {Mutiply} \\ & = & (2x \sqrt [3]{x^2-4}) \cdot \frac {(\sqrt [3]{x^2-4})^2}{(\sqrt [3]{x^2-4})^2} + \frac {2x}{(\sqrt [3]{x^2-4})^2} & \text {\small Equivalent} \\ & & & \text {\small fractions} \\ & = & \frac {2x(\sqrt [3]{x^2-4})^3}{(\sqrt [3]{x^2-4})^2} + \frac {2x}{(\sqrt [3]{x^2-4})^2} & \text {Multiply}\\ & = & \frac {2x(x^2-4)}{(\sqrt [3]{x^2-4})^2} + \frac {2x}{(\sqrt [3]{x^2-4})^2} & \text {Simplify}\\ & = & \frac {2x(x^2-4) + 2x}{(\sqrt [3]{x^2-4})^2} & \text {Add} \\ & = & \frac {2x(x^2-4 +1)}{(\sqrt [3]{x^2-4})^2} & \text {Factor}\\ & = & \frac {2x(x^2-3)}{(\sqrt [3]{x^2-4})^2} & \\ \end{array}\]
   We cannot reduce this any further because \(x^2 - 3\) is irreducible over the rational numbers.

6. We begin by working inside each set of parentheses, using the product rule for radicals and combining like terms.

   \[ \begin{array}{rclr} \sqrt {(\sqrt {18y} - \sqrt {8y})^2 + (\sqrt {20} - \sqrt {80})^2} & = & \sqrt {(\sqrt {9\cdot 2y} - \sqrt {4 \cdot 2y})^2 + (\sqrt {4\cdot 5} - \sqrt {16 \cdot 5})^2} & \\ & = & \sqrt {(\sqrt {9} \sqrt {2y} - \sqrt {4}\sqrt {2y})^2 + (\sqrt {4}\sqrt {5} - \sqrt {16}\sqrt {5})^2} & \\ & = & \sqrt {(3\sqrt {2y} - 2\sqrt {2y})^2 + (2\sqrt {5} - 4\sqrt {5})^2} & \\ & = & \sqrt {(\sqrt {2y})^2 + (-2\sqrt {5})^2} & \\ & = & \sqrt {2y + (-2)^2(\sqrt {5})^2} & \\ & = & \sqrt {2y + 4\cdot 5} & \\ & = & \sqrt {2y + 20} & \\ \end{array} \]

   To see if this simplifies any further, we factor the radicand: \(\sqrt {2y+20} = \sqrt {2(y+10)}\). Finding no perfect square factors, we are done. □

1. In our first equation, the quantity containing \(x\) is already isolated, so we extract fourth roots. The exponent is even, so when the roots are extracted we need both the positive and negative roots.

   \[ \begin{array}{rclr} (5x +3)^{4} & = & 16 & \\ 5x+3 & = & \pm \sqrt [4]{16} & \text {Extract fourth roots} \\ 5x + 3 & = & \pm 2 & \\ 5x+3 = 2 & \text {or} & 5x+3 = -2 & \\ x = -\frac {1}{5} & \text {or} & x = -1 \\ \end{array} \]
   We leave it to the reader to verify that both of these solutions satisfy the original equation.

2. In this example, we first need to isolate the quantity containing the variable \(w\). Here, third (cube) roots are required and since the exponent (index) is odd, we do not need the \(\pm \):

   \[ \begin{array}{rclr} 1 - \frac {(5-2w)^3}{7} & = & 9 & \\ - \frac {(5-2w)^3}{7} & = & 8 & \text {Subtract $1$} \\ (5-2w) ^ 3 & = & -56 & \text {Multiply by $-7$} \\ 5 - 2w & = & \sqrt [3]{-56} & \text {Extract cube root} \\ 5 - 2w & = & \sqrt [3]{(-8)(7)} & \\ 5 - 2w & = & \sqrt [3]{-8} \sqrt [3]{7} & \text {Product Rule}\\ 5 - 2w & = & -2\sqrt [3]{7} & \\ -2w & = & -5-2 \sqrt [3]{7} & \text {Subtract $5$} \\ w & = & \frac {-5 - 2\sqrt [3]{7}}{-2} & \text {Divide by $-2$} \\ w & = & \frac {5 + 2\sqrt [3]{7}}{2} & \text {Properties of Negatives} \\ \end{array}\]
   The reader should check the answer because it provides a hearty review of arithmetic.

3. To solve \(t + \sqrt {2t+3} = 6\), we first isolate the square root, then proceed to square both sides of the equation. In doing so, we run the risk of introducing extraneous solutions so checking our answers here is a necessity.

   \[ \begin{array}{rclr} t + \sqrt {2t+3} & = & 6 & \\ \sqrt {2t+3} & = & 6 - t & \text {Subtract $t$} \\ (\sqrt {2t+3})^2 & = & (6-t)^2 & \text {Square both sides} \\ 2t + 3 & = & 36-12t + t^2 & \text {F.O.I.L. / Perfect Square Trinomial} \\ 0 & = & t^2 - 14t + 33 & \text {Subtract $2t$ and $3$} \\ 0 & = & (t-3)(t-11) & \text {Factor} \\ \end{array} \]
   From the Zero Product Property, we know either \(t - 3 = 0\) (which gives \(t=3\)) or \(t-11 = 0\) (which gives \(t=11\)). When checking our answers, we find \(t = 3\) satisfies the original equation, but \(t = 11\) does not. ^[9] (It is worth noting that when \(t=11\) is substituted into the original equation, we get \(11 + \sqrt {25} = 6\). If the \(+\sqrt {25}\) were \(-\sqrt {25}\), the solution would check. Once again, when squaring both sides of an equation, we lose track of \(\pm \), which is what lets extraneous solutions in the door.) So our final answer is \(t = 3\) only.

4. In our next example, we locate the variable (in this case \(y\)) beneath a cube root, so we first isolate that root and cube both sides.

   \[ \begin{array}{rclr} \sqrt {2} - 3\sqrt [3]{2y+1} & = & 0 & \\ - 3\sqrt [3]{2y+1} & = & - \sqrt {2} & \text {Subtract $\sqrt {2}$} \\ \sqrt [3]{2y+1} & = & \frac {-\sqrt {2}}{-3} & \text {Divide by $-3$} \\ \sqrt [3]{2y+1} & = & \frac {\sqrt {2}}{3} & \text {Properties of Negatives} \\ (\sqrt [3]{2y+1})^3 & = & \left ( \frac {\sqrt {2}}{3} \right )^{3} & \text {Cube both sides} \\ 2y + 1 & = & \frac {(\sqrt {2})^3}{3^3} & \\ 2y + 1 & = & \frac {2\sqrt {2}}{27} & \\ 2y & = & \frac {2 \sqrt {2}}{27} - 1 & \text {Subtract $1$} \\ 2y & = & \frac {2 \sqrt {2}}{27} - \frac {27}{27} & \text {Common denominators} \\ 2y & = & \frac {2 \sqrt {2} - 27}{27} & \text {Subtract fractions} \\ y & = & \frac {2 \sqrt {2} - 27}{54} & \text {Divide by $2$ $\left (\text {multiply by $\frac {1}{2}$} \right )$} \\ \end{array}\]
   Since we raised both sides to an odd power, we don’t need to worry about extraneous solutions but we encourage the reader to check the solution just for the fun of it.

5. In the equation \(\sqrt {4x-1} + 2\sqrt {1 - 2x} = 1\), we have not one but two square roots. We begin by isolating one of the square roots and squaring both sides.

   \[ \begin{array}{rclr} \sqrt {4x-1} + 2\sqrt {1 - 2x} & = & 1 & \\ \sqrt {4x-1} & = & 1 - 2\sqrt {1-2x} & \text {Subtract $2\sqrt {1 - 2x}$ from both sides} \\ (\sqrt {4x-1})^2 & = & (1 - 2\sqrt {1-2x})^2 & \text {Square both sides} \\ 4x - 1 & = & 1 - 4\sqrt {1-2x} + (2\sqrt {1-2x})^2 & \text {F.O.I.L. / Perfect Square Trinomial} \\ 4x - 1 & = & 1 - 4\sqrt {1-2x} + 4(1-2x) & \\ 4x - 1 & = & 1 - 4\sqrt {1-2x} + 4 - 8x & \text {Distribute} \\ 4x - 1 & = & 5 - 8x - 4\sqrt {1-2x} & \text {Gather like terms} \\ \end{array} \]
   At this point, we have just one square root so we proceed to isolate it and square both sides a second time. ^[10] (To avoid complications with fractions, we’ll forego dividing by the coefficient of \(\sqrt {1-2x}\), namely \(-4\). This is perfectly fine so long as we don’t forget to square it when we square both sides of the equation.)
   \[ \begin{array}{rclr} 4x - 1 & = & 5 - 8x - 4\sqrt {1-2x} & \\ 12x - 6 & = & -4\sqrt {1-2x} & \text {Subtract $5$, add $8x$}\\ (12x-6)^2 & = & (-4\sqrt {1-2x})^2 & \text {Square both sides} \\ 144x^2 - 144x + 36 & = & 16(1-2x) & \\ 144x^2 - 144x + 36 & = & 16 - 32x & \\ 144x^2 - 112x + 20 & = & 0 & \text {Subtract $16$, add $32x$} \\ 4(36x^2 - 28x + 5) & = & 0 & \text {Factor} \\ 4(2x-1)(18x - 5) & = & 0 & \text {Factor some more} \\ \end{array} \]
   From the Zero Product Property, we know either \(2x-1 = 0\) or \(18x - 5 = 0\). The former gives \(x = \frac {1}{2}\) while the latter gives us \(x = \frac {5}{18}\). Since we squared both sides of the equation (twice!), we need to check for extraneous solutions. We find \(x = \frac {5}{18}\) to be extraneous, so our only solution is \(x = \frac {1}{2}\).

6. As usual, our first step in solving \(\sqrt [4]{n^2 + 2} + n = 0\) is to isolate the radical. We then proceed to raise both sides to the fourth power to eliminate the fourth root:

   \[ \begin{array}{rclr} \sqrt [4]{n^2 + 2} + n & = & 0 & \\ \sqrt [4]{n^2 + 2} & = & -n & \text {Subtract $n$} \\ (\sqrt [4]{n^2 + 2})^4 & = & (-n)^4 & \text {Raise both sides to the $4^{\text {th}}$ power} \\ n^2 + 2 & = & n^4 & \text {Properties of Negatives}\\ 0 & = & n^{4} - n^2 - 2 & \text {Subtract $n^2$ and $2$} \\ 0 & = & (n^2 - 2)(n^2 + 1) & \text {Factor - this is a `Quadratic in Disguise'} \\ \end{array} \]

   At this point, the Zero Product Property gives either \(n^2 - 2 = 0\) or \(n^2 + 1 = 0\). From \(n^2 - 2 = 0\), we get \(n^2 = 2\), so \(n = \pm \sqrt {2}\). From \(n^2 + 1 = 0\), we get \(n^2 = -1\), which gives no real solutions. ^[11] (Why is that again?) Since we raised both sides to an even (the fourth) power, we need to check for extraneous solutions. We find that \(n = -\sqrt {2}\) works but \(n = \sqrt {2}\) is extraneous.

7. In this problem, we are asked to solve for \(r\). While there are a lot of letters in this equation ^[12] (including a Greek letter, no less!) , \(r\) appears in only one term: \(r^3\). Our strategy is to isolate \(r^3\) then extract the cube root.

   \[ \begin{array}{rclr} V & = & \frac {4\pi }{3}(R^3 - r^3) & \\ 3V & = & 4\pi (R^3 - r^3) & \text {Multiply by $3$ to clear fractions}\\ 3V & = & 4\pi R^3 - 4\pi r^3 & \text {Distribute} \\ 3V - 4\pi R^3 & = & -4 \pi r^3 & \text {Subtract $4 \pi R^3$} \\ \frac {3V - 4\pi R^3}{-4\pi } & = & r^3 & \text {Divide by $-4\pi $} \\ \frac {4\pi R^3 - 3V}{4\pi } & = & r^3 & \text {Properties of Negatives} \\ \sqrt [3]{\frac {4\pi R^3 - 3V}{4\pi }} & = & r & \text {Extract the cube root} \\ \end{array} \]
   The check is, as always, left to the reader and highly encouraged.

8. The equation we are asked to solve in this example is from the world of Chemistry and is none other than Graham’s Law of Effusion. As was mentioned in Example rateqnreviewex, subscripts in Mathematics are used to distinguish between variables and have no arithmetic significance. In this example, \(r_1\), \(r_2\), \(M_1\) and \(M_2\) are as different as \(x\), \(y\), \(z\) and \(117\). Since we are asked to solve for \(M_1\), we locate \(M_1\) and see it is in the denominator of a fraction which is inside of a square root. We eliminate the square root by squaring both sides and proceed from there.

   \[ \begin{array}{rclr} \frac {r_1}{r_2} & = & \sqrt {\frac {M_2}{M_1}} & \\ \left (\frac {r_1}{r_2}\right )^2 & = & \left (\sqrt {\frac {M_2}{M_1}}\right )^2 & \text {Square both sides} \\ \frac {r_1^2}{r_2^2} & = & \frac {M_2}{M_1} & \\ r_1^2 M_1 & = & M_2r_2^2 & \text {Multiply by $r_2^2 M_1$ to clear fractions, assume $r_2$, $M_1 \neq 0$ } \\ M_1 & = & \frac {M_2r_2^2}{r_1^2} & \text {Divide by $r_1^2$, assume $r_1 \neq 0$} \\ \end{array} \]
   As the reader may expect, checking the answer amounts to a good exercise in simplifying rational and radical expressions. The fact that we are assuming all of the variables represent positive real numbers comes in to play, as well.

9. Our last equation to solve comes from Einstein’s Special Theory of Relativity and relates the mass of an object to its velocity as it moves. ^[13] (See this article on the Lorentz Factor.) We are asked to solve for \(v\) which is located in just one term, namely \(v^2\), which happens to lie in a fraction underneath a square root which is itself a denominator. We have quite a lot of work ahead of us!

   \[ \begin{array}{rclr} m & = & \frac {m_0}{\sqrt {1 - \frac {v^2}{c^2}}} & \\ m \sqrt {1 - \frac {v^2}{c^2}} & = & m_0 & \text {Multiply by $\sqrt {1 - \frac {v^2}{c^2}}$ to clear fractions}\\ \left (m \sqrt {1 - \frac {v^2}{c^2}}\right )^{2} & = & m_0^{2} & \text {Square both sides}\\ m^2 \left (1 - \frac {v^2}{c^2}\right ) & = & m_0^{2} & \text {Properties of Exponents}\\ m^2 - \frac {m^2 v^2}{c^2} & = & m_0^{2} & \text {Distribute} \\ - \frac {m^2 v^2}{c^2} & = & m_0^{2} - m^2 & \text {Subtract $m^2$} \\ m^2 v^2 & = & -c^2 (m_0^{2} - m^2) & \text {Multiply by $-c^2$ ($c^2 \neq 0$)} \\ m^2 v^2 & = & -c^2m_0^{2} + c^2 m^2 & \text {Distribute} \\ v^2 & = & \frac {c^2 m^2 -c^2m_0^{2}}{m^2} & \text {Rearrange terms, divide by $m^2$ ($m^2 \neq 0$)} \\ v & = & \sqrt {\frac {c^2 m^2 -c^2m_0^{2}}{m^2}} & \text {Extract Square Roots, $v > 0$ so no $\pm $} \\ v & = & \frac {\sqrt {c^2 (m^2 -m_0^{2})}}{\sqrt {m^2}} & \text {Properties of Radicals, factor} \\ v & = & \frac {|c|\sqrt {m^2 -m_0^{2}}}{|m|} & \\ v & = & \frac {c\sqrt {m^2 -m_0^{2}}}{m} & \text {$c > 0$ and $m > 0$ so $|c| = c$ and $|m| = m$} \\ \end{array}\]
   Checking the answer algebraically would earn the reader great honor and respect on the Algebra battlefield so it is highly recommended.

Next, we multiply the numerator and denominator by the conjugate of \(\sqrt {2x+1} - \sqrt {2x+2h+1}\), namely \(\sqrt {2x+1} + \sqrt {2x+2h+1}\), simplify and reduce:

While the denominator is quite a bit more complicated than what we started with, we have done what was asked of us. In the interest of full disclosure, the reason we did all of this was to cancel the original ‘\(h\)’ from the denominator. That’s an awful lot of effort to get rid of just one little \(h\), but you’ll see the significance of this in Calculus. □