###### Remark14.4.2

The basic strategy to solve radical equations is to isolate the radical on one side of the equation and then square both sides to cancel the radical.

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In this section, we will learn how to solve equations involving radicals.

One common application of radicals is the Pythagorean Theorem. We already saw some examples in earlier sections. We will look at some other applications of radicals in this section.

The formula \(T=2\pi\sqrt{\frac{L}{g}}\) is used to calculate the period of a pendulum and is attributed to the scientist Christiaan Huygens^{โ1โ}en.wikipedia.org/wiki/Christiaan_Huygens#Pendulums. In the formula, \(T\) stands for the pendulum's period (how long one back-and-forth oscillation takes) in seconds, \(L\) stands for the pendulum's length in meters, and \(g\) is approximately 9.8โฏ^{m}โ_{s2} which is the gravitational acceleration constant on Earth.

An engineer is designing a pendulum. Its period must be \(10\) seconds. How long should the pendulum's length be?

We will substitute \(\substitute{10}\) into the formula for \(T\) and also the value of \(g\text{,}\) and then solve for \(L\text{:}\)

\begin{align*}
T\amp=2\pi\sqrt{\frac{L}{g}}\\
\substitute{10}\amp=2\pi\sqrt{\frac{L}{\substitute{9.8}}}\\
\multiplyleft{\frac{1}{2\pi}}10\amp=\multiplyleft{\frac{1}{2\pi}}2\pi\sqrt{\frac{L}{9.8}}\\
\frac{5}{\pi}\amp=\sqrt{\frac{L}{9.8}}\\
\left(\frac{5}{\pi}\right)^\highlight{2}\amp=\left(\sqrt{\frac{L}{9.8}}\right)^\highlight{2}\amp\text{canceling square root by squaring both sides}\\
\frac{25}{\pi^2}\amp=\frac{L}{9.8}\\
\multiplyleft{9.8}\frac{25}{\pi^2}\amp=\multiplyleft{9.8}\frac{L}{9.8}\\
24.82\amp\approx L
\end{align*}

To build a pendulum with a period of \(10\) seconds, the pendulum's length should be approximately \(24.82\) meters.

The basic strategy to solve radical equations is to isolate the radical on one side of the equation and then square both sides to cancel the radical.

Squaring both sides of an equation is "dangerous", as it could create extraneous solutions, which will not make the equation true. For example, if we square both sides of \(1=-1\text{,}\) we have:

\begin{align*}
1\amp=-1\amp\text{false}\\
(1)^2\amp=(-1)^2\\
1\amp=1\amp\text{true}
\end{align*}

By squaring both sides of an equation, we turned a false equation into a true one. This is why we *must check solutions* when we square both sides of an equation.

Solve the equation \(1+\sqrt{y-1}=4\) for \(y\text{.}\)

Solution

We will isolate the radical first, and then square both sides.

\begin{align*}
1+\sqrt{y-1}\amp=4\\
\sqrt{y-1}\amp=3\\
\left(\sqrt{y-1}\right)^\highlight{2}\amp=(3)^\highlight{2}\\
y-1\amp=9\\
y\amp=10
\end{align*}

Because we squared both sides of an equation, we must check the solution. Substitute \(\substitute{10}\) into \(1+\sqrt{y-1}=4\text{,}\) and we have:

\begin{align*}
1+\sqrt{y-1}\amp=4\\
1+\sqrt{\substitute{10}-1}\amp\stackrel{?}{=}4\\
1+\sqrt{9}\amp\stackrel{?}{=}4\\
1+3\amp\stackrel{?}{=}4\\
4\amp\stackrel{\checkmark}{=}4
\end{align*}

So, \(10\) is the solution to the equation \(1+\sqrt{y-1}=4\text{.}\)

Solve the equation \(5+\sqrt{q}=3\) for \(q\text{.}\)

Solution

First, isolate the radical and square both sides.

\begin{align*}
5+\sqrt{q}\amp=3\\
\sqrt{q}\amp=-2\\
\left(\sqrt{q}\right)^\highlight{2}\amp=(-2)^\highlight{2}\\
q\amp=4
\end{align*}

Because we squared both sides of an equation, we must check the solution. Substitute \(\substitute{4}\) into \(5+\sqrt{q}=3\text{,}\) and we have:

\begin{align*}
5+\sqrt{q}\amp=3\\
5+\sqrt{\substitute{4}}\amp\stackrel{?}{=}3\\
5+2\amp\stackrel{?}{=}3\\
7\amp\stackrel{\text{no}}{=} 3
\end{align*}

Thus, the potential solution \(-2\) is actually extraneous and we have no real solutions to the equation \(5+\sqrt{q}=3\text{.}\) The solution set is the empty set, \(\emptyset\text{.}\)

In the previous example, it would be legitimate to observe earlier stages that there are no solutions. From the very beginning, how could \(5\) plus a positive quantity result in \(3\text{?}\) Or at the second step, since square roots are non-negative, how could a square root equal \(-2\text{?}\)

You do not have to be able to make these observations. If you follow the general steps for solving radical equations *and* you remember to check the possible solutions you find, then that will be enough.

Solve for \(z\) in \(\sqrt{z}+2=z\text{.}\)

Solution

We will isolate the radical first, and then square both sides.

\begin{align*}
\sqrt{z}+2\amp=z\\
\sqrt{z}\amp=z-2\\
\left(\sqrt{z}\right)^{\highlight{2}}\amp=(z-2)^{\highlight{2}}\\
z\amp=z^2-4z+4\\
0\amp=z^2-5z+4\\
0\amp=(z-1)(z-4)\\
z-1=0\amp\text{ or }z-4=0\\
z=1\amp\text{ or }z=4
\end{align*}

Because we squared both sides of an equation, we must check both solutions.

Substitute \(\substitute{1}\) into \(\sqrt{z}+2=z\text{,}\) and we have:

\begin{align*}
\sqrt{z}+2\amp=z\\
\sqrt{\substitute{1}}+2\amp\stackrel{?}{=}1\\
1+2\amp\stackrel{?}{=}1\\
3\amp\stackrel{\text{no}}{=}1
\end{align*}

It turned out \(1\) is an extraneous solution.

Next, we substitute \(\substitute{4}\) into \(\sqrt{z}+2=z\text{:}\)

\begin{align*}
\sqrt{z}+2\amp=z\\
\sqrt{\substitute{4}}+2\amp\stackrel{?}{=}4\\
2+2\amp\stackrel{?}{=}4\\
4\amp\stackrel{\checkmark}{=}4
\end{align*}

So, \(4\) is the solution.

The equation has one solution: \(4\text{.}\) The solution set is \(\{4\}\text{.}\)

Sometimes, we need to square both sides of an equation *twice* before finding the solutions, like in the next example.

Solve the equation \(\sqrt{p-5}=5-\sqrt{p}\) for \(p\text{.}\)

Solution

We cannot isolate two radicals, so we will simply square both sides, and later try to isolate the remaining radical.

\begin{align*}
\sqrt{p-5}\amp=5-\sqrt{p}\\
\left(\sqrt{p-5}\right)^\highlight{2}\amp=\left(5-\sqrt{p}\right)^\highlight{2}\\
p-5\amp=25-10\sqrt{p}+p \amp\text{ after expanding the binomial squared}\\
-5\amp=25-10\sqrt{p}\\
-30\amp=-10\sqrt{p}\\
\frac{-30}{-10}\amp=\frac{-10\sqrt{p}}{-10}\\
3\amp=\sqrt{p}\\
(3)^\highlight{2}\amp=\left(\sqrt{p}\right)^\highlight{2}\\
9\amp=p
\end{align*}

Because we squared both sides of an equation, we must check the solution by substituting \(\substitute{9}\) into \(\sqrt{p-5}=5-\sqrt{p}\text{,}\) and we have:

\begin{align*}
\sqrt{p-5}\amp=5-\sqrt{p}\\
\sqrt{\substitute{9}-5}\amp\stackrel{?}{=}5-\sqrt{9}\\
\sqrt{4}\amp\stackrel{?}{=}5-3\\
2\amp\stackrel{\checkmark}{=}2
\end{align*}

So \(9\) is the solution. The solution set is \(\{9\}\text{.}\)

Solve the equation \(\sqrt{2n-6}=1+\sqrt{n-2}\) for \(n\text{.}\)

Solution

We cannot isolate two radicals, so we will simply square both sides, and later try to isolate the remaining radical.

\begin{align*}
\sqrt{2n-6}\amp=1+\sqrt{n-2}\\
\left(\sqrt{2n-6}\right)^\highlight{2}\amp=\left(1+\sqrt{n-2}\right)^\highlight{2}\\
2n-6\amp=\left(1+\sqrt{n-2}\right)\left(1+\sqrt{n-2}\right)\\
2n-6\amp=1+\sqrt{n-2}+\sqrt{n-2}+n-2\\
2n-6\amp=2\sqrt{n-2}+n-1 \text{ by combining like terms}\\
n-5\amp=2\sqrt{n-2}\\
\end{align*}

Note here that we can leave the coefficient \(2\) next to the radical in the squaring process. We will square the \(2\) also.

\begin{align*} (n-5)^\highlight{2}\amp=\left(2\sqrt{n-2}\right)^\highlight{2}\\ n^2-10n+25\amp=4(n-2)\\ n^2-10n+25\amp=4n-8\\ n^2-14n+33\amp=0\\ (n-11)(n-3)\amp=0\\ n-11=0\amp\text{ or }n-3=0\\ n=11\amp\text{ or } n=3 \end{align*}So our two potential solutions are \(11\) and \(3\text{.}\) We should now verify that they truly are solutions.

Substitute \(\substitute{11}\) into \(\sqrt{2n-6}=1+\sqrt{n-2}\text{,}\) and we have:

\begin{align*}
\sqrt{2n-6}\amp=1+\sqrt{n-2}\\
\sqrt{2(\substitute{11})-6}\amp\stackrel{?}{=}1+\sqrt{\substitute{11}-2}\\
\sqrt{16}\amp\stackrel{?}{=}1+\sqrt{9}\\
4\amp\stackrel{?}{=}1+3\\
4\amp\stackrel{\checkmark}{=}4
\end{align*}

So, \(11\) is the solution.

Next, we substitute \(\substitute{3}\) into \(\sqrt{2n-6}=1+\sqrt{n-2} \text{,}\) and we have:

\begin{align*}
\sqrt{2n-6}\amp=1+\sqrt{n-2}\\
\sqrt{2(\substitute{3})-6}\amp\stackrel{?}{=}1+\sqrt{\substitute{3}-2}\\
\sqrt{0}\amp\stackrel{?}{=}1+\sqrt{1}\\
0\amp\stackrel{\text{no}}{=}2
\end{align*}

So, \(3\) is not a solution.

So \(11\) is the solution. The solution set is \(\{11\}\text{.}\)

We also need the ability to solve radical equations with variables, like in the next example that comes from physics. The strategy is the same: isolating the radical, and then raise both sides to a certain power to cancel the radical.

The study of black holes has resulted in some interesting science. One fundamental concept about black holes is that there is a distance close enough to the black hole that not even light can escape, called the Schwarzschild radius^{โ2โ}en.wikipedia.org/wiki/Schwarzschild_radius or the event horizon radius. To find the Schwarzschild radius, \(R_s\text{,}\) we set the formula for the escape velocity equal to the speed of light, \(c\text{,}\) and we get \(c=\sqrt{\frac{2GM}{R_s}}\) which we need to solve for \(R_s\text{.}\) Note that \(G\) is a constant, and \(M\) is the mass of the black hole.

Solution

We will start by taking the equation \(c=\sqrt{\frac{2GM}{R_s}}\) and applying our standard radical-equation-solving techniques. Isolate the radical and square both sides:

\begin{align*}
c\amp=\sqrt{\frac{2GM}{R_s}}\\
c^\highlight{2}\amp=\left(\sqrt{\frac{2GM}{R_s}}\right)^\highlight{2}\\
c^2\amp=\frac{2GM}{R_s}\\
\multiplyleft{R_s}c^2\amp=\frac{2GM}{R_s}\multiplyright{R_s}\\
R_s c^2\amp=2GM\\
\divideunder{R_s c^2}{c^2}\amp=\divideunder{2GM}{c^2}\\
R_s\amp=\frac{2GM}{c^2}
\end{align*}

So, the Schwarzschild radius can be found using the formula \(R_s=\frac{2GM}{c^2}\text{.}\)

The term redshift^{โ3โ}en.wikipedia.org/wiki/Redshift refers to the Doppler effect^{โ4โ}en.wikipedia.org/wiki/Doppler_effect for light. When an object (like a star) is moving away from Earth at very fast speeds, the wavelength of the light emitted by the star is increased due to the distance between the planets increasing (and the constant speed of light). Increased wavelength makes light โredderโ. The opposite phenomenon is called blueshift^{โ5โ}https://en.wikipedia.org/wiki/Blueshift. It turns out that the formula to calculate the redshift for a star moving away from the Earth uses square roots:

\begin{equation*}
f_r=f_s\cdot\sqrt{\frac{c-v}{c+v}}
\end{equation*}

where \(c\) stands for the constant speed of light in a vacuum, \(f_r\) represents the frequency of the light that the receiver on Earth sees, \(f_s\) represents the frequency of light that the source star emits, and \(v\) is the velocity that the star moving away from Earth. Solve this equation for \(v\text{.}\)

Solution

We will take the original equation \(f_r=f_s\cdot\sqrt{\frac{c-v}{c+v}}\) and follow the steps to solving a radical equation. We could isolate the radical and then square both sides, but in this case isolating the radical is not necessary. If we begin by squaring both sides, that too will eliminate the radical.

\begin{align*}
f_r\amp=f_s\cdot\sqrt{\frac{c-v}{c+v}}\\
\left(f_r\right)^\highlight{2}\amp=\left(f_s\cdot\sqrt{\frac{c-v}{c+v}}\right)^\highlight{2}\\
f_r^2\amp=f_s^2\cdot\frac{c-v}{c+v}\\
f_r^2\multiplyright{(c+v)}\amp=f_s^2\cdot\frac{c-v}{c+v}\multiplyright{(c+v)}\\
f_r^2(c+v)\amp=f_s^2(c-v)\\
f_r^2c+f_r^2v\amp=f_s^2c-f_s^2v\\
f_r^2v\addright{f_s^2v}\amp=f_s^2c\subtractright{f_r^2c}\\
\left(f_s^2+f_r^2\right)v\amp=\left(f_s^2-f_r^2\right)c\\
v\amp=\frac{f_s^2-f_r^2}{f_s^2+f_r^2}c
\end{align*}

This formula will tell us the velocity of the star away from Earth if we can know the respective frequencies of the starlight. This formula is used to demonstrate that the universe is expanding^{โ6โ}en.wikipedia.org/wiki/Metric_expansion_of_space.

Let's look at an example of solving an equation with a cube root. There is very little difference between solving a cube-root equation and solving a square-root equation. Instead of *squaring* both sides, you *cube* both sides.

Solve for \(q\) in \(\sqrt[3]{2-q}+2=5\text{.}\)

Solution

\begin{align*}
\sqrt[3]{2-q}+2\amp=5\\
\sqrt[3]{2-q}\amp=3\\
\left(\sqrt[3]{2-q}\right)^\highlight{3}\amp=(3)^\highlight{3}\\
2-q\amp=27\\
-q\amp=25\\
q\amp=-25
\end{align*}

Unlike squaring both sides of an equation, raising both sides of an equation to the 3rd power will not create extraneous solutions. It's still good practice to check solution, though. This part is left as exercise.

We can use technology to solve equations by finding where two graphs intersect.

Solve the equation \(1-x=\sqrt{x+5}\) with technology.

Solution
Figure14.4.14Graph of \(f(x)=1-x\) and \(g(x)=\sqrt{x+5}\)
Figure14.4.15Graph of \(f(x)=1-x\) and \(y^2=x+5\)

We define \(f(x)=1-x\) and \(g(x)=\sqrt{x+5}\text{,}\) and then look for the intersection(s) of the graphs. Since the two functions intersect at \((-1,2)\text{,}\) the solution to \(1-x=\sqrt{x+5}\) is \(-1\text{.}\) The solution set is \(\{-1\}\text{.}\)

Before we finish with this example, we would like to illustrate why there are sometimes extraneous solutions to radical equations. It has to do with the squaring-both-sides step of the solving process.

A graph of a radical, for example \(y=\sqrt{x+5}\text{,}\) actually graphs as *half* of a sideways parabola, as you can see in Figureย 14.4.14. When we square both sides of that equation, we get \(y^2=x+5\) which actually graphs as a *complete* sideways parabola.

The curve and the line now intersect *twice*! This second solution (which is \(4\text{,}\) by the way) is the extraneous solution that we would have found had we solved \(1-x=\sqrt{x+5}\) algebraically.

Solve the equation \(3+\sqrt{x+4}=x-\sqrt{x-4}\) graphically using technology.

Solution
Figure14.4.17Graph of \(m(x)=3+\sqrt{x+4}\) and \(n(x)=x-\sqrt{x-4}\)

To solve the equation graphically, first we will assign the left side of the equation the label \(m(x)=3+\sqrt{x+4}\) and the right side \(n(x)=x-\sqrt{x-4}\text{.}\) Next we will make graphs of both \(m\) and \(n\) on the same grid and look for their intersection point(s). Since the two functions intersect at about \((8.75,6.571)\text{,}\) the solution to \(3+\sqrt{x+4}=x-\sqrt{x-4}\) is \(8.75\text{.}\) The solution set is \(\{8.75\}\text{.}\)

Solving Radical Equations

Solve the equation.

\(\displaystyle{ {\sqrt{x}} = {10} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{y}} = {6} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{2y}} = {4} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{4r}} = {16} }\)

Solve the equation.

\(\displaystyle{ {3\sqrt{r}} = {6} }\)

Solve the equation.

\(\displaystyle{ {2\sqrt{t}} = {6} }\)

Solve the equation.

\(\displaystyle{ {-4\sqrt{t}} = {20} }\)

Solve the equation.

\(\displaystyle{ {-3\sqrt{t}} = {9} }\)

Solve the equation.

\(\displaystyle{ {-3\sqrt{4-x}-10} = {-40} }\)

Solve the equation.

\(\displaystyle{ {5\sqrt{-8-x}-10} = {25} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{-y+20}} = {y} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{9y+10}} = {y} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{r}+30} = {r} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{r}+6} = {r} }\)

Solve the equation.

\(\displaystyle{ {t} = {\sqrt{t+3}+9} }\)

Solve the equation.

\(\displaystyle{ {t} = {\sqrt{t-4}+6} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{t+8}} = {\sqrt{t}-4} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{x-7}} = {\sqrt{x}-1} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{x+9}} = {-1-\sqrt{x}} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{y+3}} = {-1-\sqrt{y}} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{9y}} = {8} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{6r}} = {3} }\)

Solve the equation.

\(\displaystyle{ \sqrt[3]{r-3} = {6} }\)

Solve the equation.

\(\displaystyle{ \sqrt[3]{t-9} = {10} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{5t+5}+9} = {18} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{2t+9}+6} = {15} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{x}+90} = {x} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{x}+42} = {x} }\)

Solve the equation.

\(\displaystyle{ \sqrt[3]{y-1} = {-10} }\)

Solve the equation.

\(\displaystyle{ \sqrt[3]{y-9} = {-1} }\)

Solve the equation.

\(\displaystyle{ {r} = {\sqrt{r+4}+38} }\)

Solve the equation.

\(\displaystyle{ {r} = {\sqrt{r+2}+10} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{84-t}} = {t+6} }\)

Solve the equation.

\(\displaystyle{ {\sqrt{37-t}} = {t+5} }\)

Solving Radical Equations Using Technology

Use technology to solve the equation \({\sqrt{x+0.4}}={\sqrt{x}-4.5}\text{.}\)

Use technology to solve the equation \({\sqrt{x-0.1}}={\sqrt{x}-3.1}\text{.}\)

Solving Radical Equations with Variables

Solve the equation for \(R\text{.}\) Assume that \(R\) is positive.

\(\displaystyle{{Z} = {\sqrt{L^{2}+R^{2}}}}\)

\(R =\) .

According to the Pythagorean Theorem, the length \(c\) of the hypothenuse of a rectangular triangle can be found through the following equation:

\(\displaystyle{{c} = {\sqrt{a^{2}+b^{2}}}}\)

Solve the equation for the length \(a\) of one of the triangleโs legs.

\(a =\) .

In an electric circuit, resonance occurs when the frequency \(f\text{,}\) inductance \(L\text{,}\) and capacitance \(C\) fulfill the following equation:

\(\displaystyle{{f} = {\frac{1}{2\pi \sqrt{LC}}}}\)

Solve the equation for the inductance \(L\text{.}\)

The frequency is measured in Hertz, the inductance in Henry, and the capacitance in Farad.

\(L =\) .

A pendulum has the length \(L\text{.}\) The time period \(T\) that it takes to once swing back and forth can be found with the following formula:

\(\displaystyle{{T} = {2\pi \sqrt{\frac{L}{32}}}}\)

Solve the equation for the length \(L\text{.}\)

The length is measured in feet and the time period in seconds.

\(L =\) .

Radical Equation Applications

According to the Pythagorean Theorem, the length \(c\) of the hypothenuse of a rectangular triangle can be found through the following equation:

\(\displaystyle{{c} = {\sqrt{a^{2}+b^{2}}}}\)

If a rectangular triangle has a hypothenuse of \({25\ {\rm ft}}\) and one leg is \({24\ {\rm ft}}\) long, how long is the third side of the triangle?

The third side of the triangle is long.

According to the Pythagorean Theorem, the length \(c\) of the hypothenuse of a rectangular triangle can be found through the following equation:

\(\displaystyle{{c} = {\sqrt{a^{2}+b^{2}}}}\)

If a rectangular triangle has a hypothenuse of \({25\ {\rm ft}}\) and one leg is \({24\ {\rm ft}}\) long, how long is the third side of the triangle?

The third side of the triangle is long.

In a coordinate system, the distance \(r\) of a point \((x,y)\) from the origin \((0,0)\) is given by the following equation:

\(\displaystyle{{r} = {\sqrt{x^{2}+y^{2}}}}\)

If a point in a coordinate system is \({41\ {\rm cm}}\) away from the origin and its x coordinate is \({40\ {\rm cm}}\text{,}\) what is its \(y\) coordinate? Assume that \(y\) is positive.

\(y =\) .

In a coordinate system, the distance \(r\) of a point \((x,y)\) from the origin \((0,0)\) is given by the following equation:

\(\displaystyle{{r} = {\sqrt{x^{2}+y^{2}}}}\)

If a point in a coordinate system is \({41\ {\rm cm}}\) away from the origin and its x coordinate is \({40\ {\rm cm}}\text{,}\) what is its \(y\) coordinate? Assume that \(y\) is positive.

\(y =\) .

A pendulum has the length \(L\) ft. The time period \(T\) that it takes to once swing back and forth is \(2\) s. Use the following formula to find its length:

\(\displaystyle{{T} = {2\pi \sqrt{\frac{L}{32}}}}\)

The pendulum is long.

A pendulum has the length \(L\) ft. The time period \(T\) that it takes to once swing back and forth is \(2\) s. Use the following formula to find its length:

\(\displaystyle{{T} = {2\pi \sqrt{\frac{L}{32}}}}\)

The pendulum is long.