6.3: Radical Equations
- Page ID
- 197602
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As is the theme in College Algebra, we begin this section with the assumption that you already have exposure to solving simple radical equations. We still spend a few moments recalling old techniques, but quickly jump into solving more complex radical equations.
Solving Radical Equations Containing One Radical
We start our trip down memory lane by recalling an old definition.
When solving these equations, our goal is to isolate the radical. At that point, our strategy will be to raise both sides of the equation to the power of the index. This will, in turn, eliminate the radical; however, we have something to watch out for when solving radical equations!
If, while solving a radical equation, we obtain a "solution" that would otherwise cause an even-indexed radical to be negative or the argument of the radical to be negative, then we know this "solution" will not work. In such cases we call this an extraneous solution as we did when we solved rational equations.
As the definition suggests, extraneous solutions are always a concern when functions within an equation have implicit domain restrictions (e.g., rational and radical functions).
In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index \(n\) to the \(n^{\text{th}}\) power. This will eliminate the radical.
Solve: \(\sqrt{5 n-4}-9=0\).
- Solution
- \[ \begin{array}{rrclcl}
& \sqrt{5n - 4} - 9 & = & 0 & & \\[6pt]
\implies & \sqrt{5n - 4} & = & 9 & \quad & \left( \text{isolating the radical by adding }9\text{ to both sides} \right) \\[6pt]
\implies & \left(\sqrt{5n - 4}\right)^2 & = & 9^2 & \quad & \left( \text{squaring both sides. }\textbf{Warning: }\text{This forces us to check our final solution} \right) \\[6pt]
\implies & 5n - 4 & = & 81 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 5n & = & 85 & \quad & \left( \text{adding }4\text{ to both sides} \right) \\[6pt]
\implies & n & = & 17 & \quad & \left( \text{dividing both sides by 5} \right) \\[6pt]
\end{array} \nonumber \]Now we must to check the solution. When doing so, you need to go back to the original equation to check your answer.\[ \sqrt{5(17) - 4} - 9 = \sqrt{85 - 4} - 9 = \sqrt{81} - 9 = 9 - 9 = 0. \checkmark \nonumber \]Thus, \( n = 17 \) is the solution to the equation.
Solve: \(\sqrt{3 m+2}-5=0\).
- Answer
-
\(m=\frac{23}{3}\)
It is important to remember the following advice:
Whenever you raise both sides of an equation to an even power, you must check your final solutions.
In truth, the reason why we (as mathematicians) ask you to do this is to catch any inconsistencies that might have slipped past you. For example, suppose we were asked to solve \( \sqrt{3x - 1} - 7 = -9 \). We would start by isolating the radical to get \( \sqrt{3x - 1} = -2 \). If you are paying attention (and not asleep at the "mathematical wheel"), you would notice that you have an even-indexed radical equaling a negative number - this cannot happen! At this point, you would stop the solution process and state that the solution doesn't exist (some people use the notation "DNE" for "does not exist").
However, if you didn't notice the negative on the right side, you would continue to solve by squaring both sides to get\[ 3x - 1 = 4 \implies 3x = 5 \implies x = \dfrac{5}{3}. \nonumber \]As long as we develop the habit to check our solutions whenever we square both sides of an equation, we have a second safety net.\[ \sqrt{3 \left( \dfrac{5}{3} \right) - 1} - 7 = \sqrt{5 - 7} - 7 = \sqrt{-2} - 7 \nonumber \]which is not a real number. As such, it cannot become \( -9 \).
I often try to get my students to stop at each step of the solution process to think about whether or not the statement before them makes sense, but checking your work at the end is your second opportunity to catch the logical flaw.
Solve: \(\sqrt{9 k-2}+1=0\).
- Solution
- \[ \begin{array}{rrclcl}
& \sqrt{9k - 2} + 1 & = & 0 & & \\[6pt]
\implies & \sqrt{9k - 2} & = & -1 & \quad & \left( \text{isolating the radical by subtracting }1\text{ from both sides} \right) \\[6pt]
\end{array} \nonumber \]At this point, if we were paying attention, we could stop because a square root is never equal to \( -1 \) (at least, not over the real numbers).
Having recognized the square root equaling a negative number in the last step of Example \( \PageIndex{ 2 } \) saved us time. Had we not seen that issue, we would have continued on to get\[ \begin{array}{rrclcl}
& \sqrt{9k - 2} + 1 & = & 0 & & \\[6pt]
\implies & \sqrt{9k - 2} & = & -1 & \quad & \left( \text{isolating the radical by subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & 9k - 2 & = & 1 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & 9k & = & 3 & \quad & \left( \text{adding }2\text{ to both sides} \right) \\[6pt]
\implies & k & = & \dfrac{1}{3} & \quad & \left( \text{dividing both sides by }9 \right) \\[6pt]
\end{array} \nonumber \]Because we squared both sides, we go back to the original equation and check our answer.\[ \sqrt{9\left( \dfrac{1}{3} \right) - 2} + 1 = \sqrt{3 - 2} + 1 = \sqrt{1} + 1 = 1 + 1 = 2 \neq 0. \nonumber \]This is a contradiction. Hence, \( k = \frac{1}{3} \) is an extraneous solution.
While we are at it, I might as well throw in the definition of contradiction for the curious mind.
Solve: \(\sqrt{2 r-3}+5=0\).
- Answer
-
no solution
Solve: \(\sqrt{p-1}+1=p\).
- Solution
- \[ \begin{array}{rrclcl}
& \sqrt{p - 1} + 1 & = & p & & \\[6pt]
\implies & \sqrt{p - 1} & = & p - 1 & \quad & \left( \text{isolating the radical by subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & \left(\sqrt{p - 1}\right)^2 & = & \left(p - 1\right)^2 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & p - 1 & = & p^2 - 2p + 1 & \quad & \left( \text{simplifying and distributing} \right) \\[6pt]
\implies & 0 & = & p^2 - 3p + 2 & \quad & \left( \text{subtracting }p\text{ and adding }1\text{ to both sides} \right) \\[6pt]
\implies & 0 & = & (p - 2)(p - 1) & \quad & \left( \text{factoring} \right) \\[6pt]
\end{array} \nonumber \]Therefore, our candidate solutions are \( p = 2 \) or \( p = 1 \). However, because we squared both sides, we need to check these results (using the original equation).\[ \begin{array}{rrcl|rrcl}
p = 2 & & & & p = 1 & & & \\[6pt]
\implies & \sqrt{2 - 1} + 1 & \overset{\text{?}}{=} & 2 & \implies & \sqrt{1 - 1} + 1 & \overset{\text{?}}{=} & 1 \\[6pt]
\implies & \sqrt{1} + 1 & \overset{\text{?}}{=} & 2 & \implies & \sqrt{0} + 1 & \overset{\text{?}}{=} & 1 \\[6pt]
\implies & 1 + 1 & \overset{\text{?}}{=} & 2 & \implies & 0 + 1 & \overset{\text{?}}{=} & 1 \\[6pt]
\implies & 2 & = & 2 \checkmark & \implies & 1 & = & 1 \checkmark \\[6pt]
\end{array} \nonumber \]Therefore, the solutions are \(p=1\) and \(p=2\).
Solve: \(\sqrt{x-2}+2=x\).
- Answer
-
\(x=2, x=3\)
It should be obvious that when the index of the radical is \(n\), we eventually raise both sides of the radical equation to the \( n^{\text{th}} \) power.
Solve: \(\sqrt[3]{5 x+1}+8=4\).
- Solution
- \[ \begin{array}{rrclcl}
& \sqrt[3]{5x + 1} + 8 & = & 4 & & \\[6pt]
\implies & \sqrt[3]{5x + 1} & = & -4 & \quad & \left( \text{isolating the radical by subtracting }4\text{ from both sides} \right) \\[6pt]
\implies & \left(\sqrt[3]{5x + 1}\right)^3 & = & \left(-4\right)^3 & \quad & \left( \text{raising both sides to the third power} \right) \\[6pt]
\implies & 5x + 1 & = & -64 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 5x & = & -65 & \quad & \left( \text{subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & x & = & -13 & \quad & \left( \text{dividing both sides by }5 \right) \\[6pt]
\end{array} \nonumber \]
It's always a good idea to check your answers when solving equations in Mathematics; however, it's only truly required (at least for now) when we have raised both sides of an equation to an even power (e.g., when solving radical equations) or when we have multiplied both sides of an equation by an expression containing a variable (e.g., when solving a rational equation). This is why we did not bother checking our solution in Example \( \PageIndex{ 4 } \). We leave it to the reader to check that \( x = -13 \) works.
Solve: \( \sqrt[3]{4 x-3}+8=5\)
- Answer
-
\(x=-6\)
Solving Equations Containing Rational Exponents
Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical equation. Since \( a^{m/n} = \sqrt[n]{a^m} \), we first raise each side of the equation to the power of the denominator of the rational exponent. Once that is done, we take the \( m^{\text{th}} \) root of both sides of the equation (or raise both sides to the \( \frac{1}{m}^{\text{th}} \) power).
Solve: \((3 x-2)^{1/4}+3=5\).
- Solution
- \[ \begin{array}{rrclcl}
& (3 x-2)^{1/4}+3 & = & 5 & & \\[6pt]
\implies & (3 x-2)^{1/4} & = & -2 & \quad & \left( \text{subtracting }3\text{ from both sides} \right) \\[6pt]
\implies & \left((3 x-2)^{1/4}\right)^4 & = & \left(-2\right)^4 & \quad & \left( \text{raising both sides to the fourth power} \right) \\[6pt]
\implies & (3 x-2)^{4/4} & = & 16 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & (3 x-2)^{1} & = & 16 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 3 x-2 & = & 16 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 3 x & = & 18 & \quad & \left( \text{adding }2\text{ to both sides} \right) \\[6pt]
\implies & x & = & 6 & \quad & \left( \text{dividing both sides by }3 \right) \\[6pt]
\end{array} \nonumber \]Since we raised both sides to an even power, we need to check for extraneous solutions. Going back to the original equation and substituting \( x = 6 \), we get\[ \left( 3(6) - 2 \right)^{1/4} + 3 = \left( 18 - 2 \right)^{1/4} + 3 = (16)^{1/4} + 3 = \sqrt[4]{16} + 3 = 2 + 3 = 5 \, \checkmark. \nonumber \]
Solve: \((9 x+9)^{1/4}-2=1\)
- Answer
-
\(x=8\)
As with radical equations, care must be taken when solving equations containing rational exponents. We must remember all of the techniques we previously learned. For example, in the equation \( a^{4/7} = 1\), we start by raising both sides to the \( 7^{\text{th}} \) power to get\[ a^4 = 1^7 = 1. \nonumber \]However, we now have the variable \( a \) raised to an even power. Extraction of Roots tells us that \( a = \pm 1 \). This is important!
Solve: \( 3\left( \frac{1}{2}x + 1 \right)^{2/3} - 5 = 7 \)
- Solution
- \[ \begin{array}{rrclcl}
& 3\left( \dfrac{1}{2}x + 1 \right)^{2/3} - 5 & = & 7 & & \\[6pt]
\implies & 3\left( \dfrac{1}{2}x + 1 \right)^{2/3} & = & 12 & \quad & \left( \text{begin isolating the rational exponent expression by adding }5\text{ to both sides} \right) \\[6pt]
\implies & \left( \dfrac{1}{2}x + 1 \right)^{2/3} & = & 4 & \quad & \left( \text{continue isolating the rational exponent expression by dividing both sides by }3 \right) \\[6pt]
\implies & \left(\left( \dfrac{1}{2}x + 1 \right)^{2/3}\right)^3 & = & \left(4\right)^3 & \quad & \left( \text{raise both sides to the third power} \right) \\[6pt]
\implies & \left( \dfrac{1}{2}x + 1 \right)^{6/3} & = & 64 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & \left( \dfrac{1}{2}x + 1 \right)^{2} & = & 64 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & \dfrac{1}{2}x + 1 & = & \pm 8 & \quad & \left( \text{Extraction of Roots} \right) \\[6pt]
\implies & \dfrac{1}{2}x & = & -1 \pm 8 & \quad & \left( \text{subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & x & = & -2 \pm 16 & \quad & \left( \text{multiplying both sides by }2 \right) \\[6pt]
\end{array} \nonumber \]Thus, the solutions are \( x = -2 - 16 = -18 \) and \( x = -2 + 16 = 14 \).
Solve: \( 6\left( 3x - 5 \right)^{3/2} - 4 = 2 \)
- Answer
-
\( x = 2 \)
Solving Radical Equations with Two Radicals
If we have a radical equation containing two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first. In the next example, when one radical is isolated, the second radical is also isolated.
Solve: \(\sqrt[3]{4 x-3} - \sqrt[3]{3 x+2} = 0\).
- Solution
- \[ \begin{array}{rrclcl}
& \sqrt[3]{4 x-3} - \sqrt[3]{3 x+2} & = & 0 & & \\[6pt]
\implies & \sqrt[3]{4 x-3} & = & \sqrt[3]{3 x+2} & \quad & \left( \text{isolating one of the radicals (which actually isolates the other in this example)} \right) \\[6pt]
\implies & \left(\sqrt[3]{4 x-3}\right)^3 & = & \left(\sqrt[3]{3 x+2}\right)^3 & \quad & \left( \text{raising both sides to the third power} \right) \\[6pt]
\implies & 4 x-3 & = & 3 x+2 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & x & = & 5 & \quad & \left( \text{subtracting }3x\text{ and adding }3\text{ to both sides} \right) \\[6pt]
\end{array} \nonumber \]The solution is \(x=5\). We leave it to you to show that \(5\) checks.
Solve: \(\sqrt[3]{7 x+1}=\sqrt[3]{2 x-5}\).
- Answer
-
\(x=-\frac{6}{5}\)
Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we "rinse and repeat" by isolating the remaining radical and raising both sides of the equation to the power of the index again.
Solve: \(\sqrt{m}+1=\sqrt{m+9}\).
- Solution
- \[ \begin{array}{rrclcl}
& \sqrt{m} + 1 & = & \sqrt{m + 9} & & \\[6pt]
\implies & \left( \sqrt{m} + 1 \right)^2 & = & \left( \sqrt{m + 9} \right)^2 & \quad & \left( \text{since a radical was already isolated, square both sides} \right) \\[6pt]
\implies & \left( \sqrt{m} + 1 \right)\left( \sqrt{m} + 1 \right) & = & m + 9 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & m + 2 \sqrt{m} + 1 & = & m + 9 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & 2 \sqrt{m} & = & 8 & \quad & \left( \text{begin isolating the remaining radical by subtracting both }m\text{ and }1\text{ from both sides} \right) \\[6pt]
\implies & \sqrt{m} & = & 4 & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt]
\implies & \left(\sqrt{m}\right)^2 & = & 4^2 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & m & = & 16 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Since we squared both sides during our solution process, we need to check that our solution works in the original equation.\[ \begin{array}{rrcl}
& \sqrt{16} + 1 & \overset{\text{?}}{=} & \sqrt{16 + 9} \\[6pt]
\implies & 4 + 1 & \overset{\text{?}}{=} & \sqrt{25} \\[6pt]
\implies & 5 & = & 5 \, \checkmark \\[6pt]
\end{array} \nonumber \]Therefore, the solution is \( m = 16 \).
Solve: \(3-\sqrt{x}=\sqrt{x-3}\).
- Answer
-
\(x=4\)
We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation. This procedure will now work for any radical equation.
Be careful as you square binomials (as in in the next example). Remember the pattern in \((a+b)^{2}=a^{2}+2 a b+b^{2}\) or \((a-b)^{2}=a^{2}-2 a b+b^{2}\).
Solve: \(\sqrt{q-2}+3=\sqrt{4 q+1}\).
- Solution
- \[ \begin{array}{rrclcl}
& \sqrt{q-2} + 3 & = & \sqrt{4q+1} & & \\[6pt]
\implies & \left( \sqrt{q-2}+3 \right)^2 & = & \left( \sqrt{4 q+1} \right)^2 & \quad & \left( \text{squaring both sides since one radical is already isolated} \right) \\[6pt]
\implies & \left( \sqrt{q-2}+3 \right)\left( \sqrt{q-2}+3 \right) & = & 4 q+1 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & q - 2 + 6 \sqrt{q-2} + 9 & = & 4 q+1 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & q + 7 + 6 \sqrt{q-2} & = & 4 q+1 & \quad & \left( \text{combining like terms} \right) \\[6pt]
\implies & 6 \sqrt{q-2} & = & 3 q - 6 & \quad & \left( \text{begin isolating the remaining radical by subtracting }q\text{ and }7\text{ from both sides} \right) \\[6pt]
\implies & 2 \sqrt{q-2} & = & q - 2 & \quad & \left( \text{noticing that all coefficients are share a common divisor, we divide both sides by }3 \right) \\[6pt]
\implies & \left(2 \sqrt{q-2}\right)^2 & = & \left(q - 2\right)^2 & \quad & \left( \text{instead of dividing by }2\text{ (which would introduce fractions), we square both sides} \right) \\[6pt]
\implies & 2^2 \left(\sqrt{q-2}\right)^2 & = & \left(q - 2\right)\left(q - 2\right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & 4 \left(q-2\right) & = & q^2 - 4q + 4 & \quad & \left( \text{simplifying and distributing} \right) \\[6pt]
\implies & 4 q - 8 & = & q^2 - 4q + 4 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & 0 & = & q^2 - 8q + 12 & \quad & \left( \text{subtracting }4q\text{ and adding }8\text{ to both sides} \right) \\[6pt]
\implies & 0 & = & \left(q - 6\right)\left(q - 2\right) & \quad & \left( \text{factoring the trinomial} \right) \\[6pt]
\end{array} \nonumber \]Therefore, our candidate solutions are \( q = 6 \) and \( q = 2 \). We leave it to the reader to check that neither is extraneous.
Solve: \(\sqrt{x-1}+2=\sqrt{2 x+6}\)
- Answer
-
\(x=5\)
Equations involving rational exponents are fertile territory to run into equations that are quadratic-in-form. Not only must you recognize their form, but you also must watch out for all the pitfalls mentioned throughout this section concerning checking your solutions.
Solve: \( x - 2 \sqrt{x} - 8 = 0 \).
- Solution
- This is a radical equation that can be written in quadratic form. If we let \( u = \sqrt{x} \), then \( u^2 = \left( \sqrt{x} \right)^2 = x \) and we can write\[ \begin{array}{rrclcl}
& x - 2\sqrt{x} - 8 & = & 0 & & \\[6pt]
\implies & u^2 - 2u - 8 & = & 0 & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & (u - 4)(u + 2) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt]
\end{array} \nonumber \]Therefore, \( u = 4 \) or \( u = -2 \). Since \( u = \sqrt{x} \), we resubstitute to get \( \sqrt{x} = 4 \) or \( \sqrt{x} = -2 \); however, square roots are never negative. Hence, we throw out that possibility and focus on \( \sqrt{x} = 4 \implies x = 16 \). Checking this solution in the original equation, we get\[ 16 - 2 \sqrt{16} - 8 = 16 - 2(4) - 8 = 16 - 8 - 8 = 0. \, \checkmark \nonumber \]Thus, the solution to the equation is \( x = 16 \).
Solve: \( \sqrt{x} - \sqrt[4]{x} - 2 = 0 \)
- Answer
-
\( x = 16 \)
While Example \( \PageIndex{ 10 } \) is interesting, the following example showcases a more frequent occurrence.
Solve: \( x^{-2/3} - 3x^{-1/3} - 10 = 0 \).
- Solution
- If we let \( u = x^{-1/3} \), then \( u^2 = \left( x^{-1/3} \right)^2 = x^{-2/3} \) and we can write\[ \begin{array}{rrclcl}
& x^{-2/3} - 3x^{-1/3} - 10 & = & 0 & & \\[6pt]
\implies & u^2 - 3u - 10 & = & 0 & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & (u - 5)(u + 2) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt]
\implies & \left(x^{-1/3} - 5\right)\left(x^{-1/3} + 2\right) & = & 0 & \quad & \left( \text{resubstituting} \right) \\[6pt]
\end{array} \nonumber \]Therefore, \( x^{-1/3} = 5 \) or \( x^{-1/3} = -2 \).\[ \begin{array}{rrcl|rclcl}
& x^{-1/3} & = & 5 & x^{-1/3} & = & -2 & & \\[6pt]
\implies & \dfrac{1}{x^{1/3}} & = & 5 & \dfrac{1}{x^{1/3}} & = & -2 & \quad & \left( \text{Laws of Exponents} \right)\\[6pt]
\implies & x^{1/3} & = & \dfrac{1}{5} & x^{1/3} & = & -\dfrac{1}{2} & \quad & \left( \text{if }a=b,\text{ where neitehr are zero, then }\frac{1}{a} = \frac{1}{b} \right) \\[6pt]
\implies & \left(x^{1/3}\right)^3 & = & \left(\dfrac{1}{5}\right)^3 & \left(x^{1/3}\right)^3 & = & \left(-\dfrac{1}{2}\right)^3 & \quad & \left( \text{raising both sides to the third power} \right) \\[6pt]
\implies & x^{3/3} & = & \dfrac{1^3}{5^3} & x^{3/3} & = & -\dfrac{1^3}{2^3} & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & x & = & \dfrac{1}{125} & x & = & -\dfrac{1}{8} & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]The domain of the original equation only has the restriction that \( x \neq 0 \) (can you see why?). Furthermore, the solution process involved cubing, not squaring. Squaring can introduce extraneous solutions, but cubing does not, so a check is not necessary. We leave it to the reader to check that these solutions work in the original equation.
Applications Involving Radicals
As you progress through your college courses, you'll encounter formulas that include radicals in many disciplines. The following Problem Solving Strategy is incredibly helpful as you progress through College Algebra and into Calculus.
- Read the problem carefully and get clarification on words or concepts that are foreign to you.
- Understand what is being stated and asked by rephrasing statements into language you connect with. Draw a figure to represent the situation, if possible.
- Label the given and wanted information. The wanted information should have a variable representing it.
- Translate into an equation relating the given and wanted information by writing the appropriate formula or model for the situation. Substitute in the given information.
- Solve the equation using good Algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- State the answer to the question with a complete sentence.
One application of radicals has to do with the effect of gravity on falling objects. The following (non-memorizable) formula allows us to determine how long it will take a fallen object to hit the ground.
On Earth, if an object is dropped from a height of \(h\) feet (where \( h \) is not too large) and if we ignore air resistance, the time (in seconds) it will take to reach the ground is approximated by using the formula \( t= \frac{\sqrt{h}}{4} \).
For example, if an object is dropped from a height of \(64\) feet, we can find the time it takes to reach the ground by substituting \(h=64\) into the formula.\[ \begin{array}{rrclcl}
& t & = & \dfrac{\sqrt{h}}{4} & & \\[6pt]
\implies & t & = & \dfrac{\sqrt{64}}{4} & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & t & = & \dfrac{8}{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & t & = & 2 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Therefore, it would take \(2\) seconds for an object dropped from a height of \(64\) feet to reach the ground (ignoring deceleration from air resistance).
The previous theorem is (for most instructors) a non-memorizable theorem. It's just a shortcut to approximate the time it takes for an object to fall a certain distance (and, in truth, it is only partially accurate for short falls).
Kaitlyn dropped her sunglasses from a bridge \(400\) feet above a river. How many seconds does it take for the sunglasses to reach the river?
- Solution
-
- Read the problem.
- Understand the situation. We are looking for the time it takes for Kaitlyn's sunglasses to reach the river.
- Label the given and wanted information. Let \( t \) represent the unknown time it will take the sunglasses to fall to the river.\[ \begin{array}{c|c}
\textbf{Given} & \textbf{Want} \\[6pt]
\hline h = 400 & t \\[6pt]
\end{array} \nonumber \] - Translate into an equation. The recent theorem allows us to relate \( h \) and \( t \) by the formula\[ t = \dfrac{\sqrt{h}}{4}. \nonumber \]Substituting in the given information, we get\[ t = \dfrac{\sqrt{400}}{4}. \nonumber \]
- Solve the equation.\[ \begin{array}{rclcl}
t & = & \dfrac{\sqrt{400}}{4} & & \\[6pt]
& = & \dfrac{20}{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 5 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \] - Check the solution.\[ \begin{array}{rrcl}
& 5 & \overset{\text{?}}{=} & \dfrac{\sqrt{400}}{4} \\[6pt]
\implies & 5 & \overset{\text{?}}{=} & \dfrac{20}{4} \\[6pt]
\implies & 5 & = & 5 \, \checkmark \\[6pt]
\end{array} \nonumber \]Thus, \( 5 \) is correct mathematically. It also makes sense logically (for example, if the answer was negative, that would not make sense). - State the answer. It takes \( 5 \) seconds for Kaitlyn's sunglasses to reach the river.
A window washer dropped a squeegee from a platform \(196\) feet above the sidewalk. How many seconds does it take for the squeegee to reach the sidewalk?
- Answer
-
\(3.5\) seconds
Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.
If the length of the skid marks is \(d\) feet, then the speed, \(s\), of the car (in miles per hour) before the brakes were applied can be found by using the formula \(s=\sqrt{24 d}\).
Again, this formula is not memorizable and it is only accurate under special circumstances (e.g., it is not accurate at all if the roadway was covered in ice).
After a car accident, the skid marks for one car measured \(190\) feet. Find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.
- Solution
-
- Read!
- Understand: We are looking for the initial speed of the car before the brakes were applied.
- Label: Let \( s \) represent the unknown speed of the car and \( d \) represent the length of the skid.\[ \begin{array}{c|c}
\textbf{Given} & \textbf{Want} \\[6pt]
\hline d = 190 & s \\[6pt]
\end{array} \nonumber \] - Translate:\[ s = \sqrt{24 d} = \sqrt{24(190)}. \nonumber \]
- Solve:\[ \begin{array}{rclcl}
s & = & \sqrt{24(190)} & & \\[6pt]
& = & \sqrt{4560} & \quad & \left( \text{simplifying} \right) \\[6pt]
& \approx & 67.5277... & \quad & \left( \text{approximating} \right) \\[6pt]
& \approx & 67.5 & \quad & \left( \text{rounding to the nearest tenth} \right) \\[6pt]
\end{array} \nonumber \] - Check:\[ \begin{array}{rrcl}
& 67.5 & \overset{\text{?}}{\approx} & \sqrt{24(190)} \\[6pt]
\implies & 67.5 & \overset{\text{?}}{\approx} & \sqrt{4560} \\[6pt]
\implies & 67.5 & \approx & 67.5277... \, \checkmark \\[6pt]
\end{array} \nonumber \]Thus, \( 67.5 \) is correct mathematically. It also makes sense logically that the car could have been going \( 67.5 \) miles per hour. - State: The speed of the car before the brakes were applied was roughly \(67.5\) miles per hour.
An accident investigator measured the skid marks of the car. The length of the skid marks was \(76\) feet. Find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.
- Answer
-
\(42.7\) mph