## 8.1 Design a Pendulum clock

A pendulum clock is a clock that uses a pendulum, a swinging weight, as its timekeeping element. Galileo Galilei discovered in early 17th century the relation between the length $$L$$ of a pendulum and the period $$T$$ of th pendulum. For a pendulum clock, the relations is approximately determined by the following rule of thumb formula: $T\approx 2\sqrt{L}$ given that $$L$$ and $$T$$ are measured in meters and seconds respectively. If the period of a pendulum clock is 2 seconds, how long should be the pendulum?

## 8.2 Solving Radical Equations by Taking a Power

The idea to solve a radical equation $$\sqrt[n]{X}=a$$ is to first take $$n$$-th power of both sides to get rid of the radical sign, that is $$X=a^n$$ and then solve the resulting equation.

Solve by Reduction
The goal to solve a single variable equation is to isolate the variable. When an equation involves radical expressions, you can not isolate the variable arithmetically without eliminating the radical sign unless the radicand is a perfect power. To remove a radical sign, you make take a power. However, youâ€™d better to isolate it first. Because simply taking powers of both sides may create new radical expressions.

Example 8.1 Solve the equation $$x-\sqrt{x+1}=1.$$

Solution.

1. Arrange terms so that one radical is isolated on one side of the equation. $x-1=\sqrt{x+1}$

2. Square both sides to eliminate the square root. $(x-1)^2=x+1$

3. Solve the resulting equation. \begin{aligned} x^2-2x+1&=x+1\\ x^2-3x&=0\\ x(x-3)&=0 \end{aligned} \begin{aligned} x =0 & \qquad \text{or} & x-3 =0 \\ x =0 & \qquad \text{or} & x =3 \end{aligned}

4. Check all proposed solutions.
Plug $$x=0$$ into the original equation, we see that the left hand side is $$0-\sqrt{0+1}=0-\sqrt{1}=0-1=-1$$ which is not equal to the right hand side. So $$x=0$$ cannot be a solution.

Plug $$x=3$$ into the original equation, we see that the left hand side is $$3-\sqrt{3+1}=3-\sqrt{4}=3-2=1$$. So $$x=3$$ is a solution.

Example 8.2 Solve the equation $$\sqrt{x-1}-\sqrt{x-6}=1.$$

Solution.

1. Isolated one radical. $\sqrt{x-1}=\sqrt{x-6}+1\\$
2. Square both sides to remove radical sign and then isolate the remaining radical. \begin{aligned} x-1&=(x-6)+2\sqrt{x-6}+1\\ x-1&=x-5+2\sqrt{x-6}\\ 4&=2\sqrt{x-6}\\ 2&=\sqrt{x-6}. \end{aligned}
3. Square both sides to remove the radical sign and then solve. \begin{aligned} \sqrt{x-6}&=2\\ x-6&=4\\ x&=10. \end{aligned} Since $$10-1>0$$ and $$10-6>0$$, $$x=10$$ is a valid solution. Indeed, $\sqrt{10-1}-\sqrt{10-6}=\sqrt{9}-\sqrt{4}=3-2=1.$

Example 8.3 Solve the equation $$-2\sqrt[3]{x-4}=6.$$

Solution.

1. Isolated the radical. $\sqrt[3]{x-4}=-3$
2. Cube both sides to eliminate the cube root and then solve the resulting equation. \begin{aligned} x-4&=(-3)^3\\ x-4&=-27\\ x&=-23 \end{aligned} The solution is $$x=-23$$.

## 8.3 Equations Involving Rational Exponents

Equation involving rational exponents may be solved similarly. However, one should be careful with meaning of the expression $$\left(X^{\frac mn}\right)^{\frac nm}$$. When $$m$$ is even and $$n$$ is odd, $$\left(X^{\frac mn}\right)^{\frac nm}=|X|$$. Otherwise, $$\left(X^{\frac mn}\right)^{\frac nm}=X$$.

Example 8.4 Solve the equation $$(x+2)^{\frac12}-(x-3)^{\frac12}=1$$.

Solution.

Since there are more than one term involving rational exponents, to solve the equation, we isolate one term and taking power and so on so forth. \begin{aligned} (x+2)^{\frac12}-(x-3)^{\frac12}=&1\\ (x+2)^{\frac12}=& (x-3)^{\frac12}+1\\ x+2 =& \left((x-3)^{\frac12}+1\right)^2\\ x+2 =& (x-3)+2(x-3)^{\frac12}+1\\ 2(x-3)^{\frac12}=&4\\ (x-3)^{\frac12}=&2\\ x-3=&4\\ x=&7 \end{aligned} Check: $(7+2)^{\frac12}-(7-3)^{\frac12}=\sqrt{9}-\sqrt{4}=3-2=1.$

So the equation has one solution $$x=7$$.

Example 8.5 Solve the equation $$(x-1)^{\frac{2}{3}}=4$$.

Solution.

There are different way to solve this equation. One may is to take rational powers of both sides and solve the resulting equation. \begin{aligned} (x-1)^{\frac{2}{3}}=&4\\ \left((x-1)^{\frac{2}{3}}\right)^{\frac32}=&4^{\frac32}\\ |x-1|=&8\\ x-1=8 \qquad\text{or}&\qquad x-1=-8\\ x=9 \qquad\text{or}&\qquad x=-7 \end{aligned} Check: $(9-1)^{\frac23}=8^{\frac23}=(8^{\frac13})^2=2^2=4;$ $(-7-1)^{\frac23}=(-8)^{\frac23}=((-8)^{\frac13})^2=(-2)^2=4.$

So the equation has two solutions $$x=9$$ and $$x=-7$$.

## 8.4 Learn from Mistakes

Example 8.6 Can you find the mistakes made in the solution and fix it?

Solve the radical equation. $\sqrt{x-1}+2=x$

Solution (incorrect): \begin{aligned} \sqrt{x-2}+2&=x\\ (\sqrt{x-2})^2+2^2&=x^2\\ x-2+4&=x^2\\ x+2&=x^2\\ x^2-x-2&=0\\ (x-2)(x+1)&=0\\ x-2=0 \qquad\text{or}& \qquad x+1=0\\ x=2 \qquad\text{or}&\qquad x=-1 \end{aligned} Answer: the equation has two solutions $$x=2$$ and $$x=-1$$.

Solution.

When squaring one side of the equation, the other side as a whole should be squared. The mistake occurred at the squaring step. The right way to solve the equation is as follows. \begin{aligned} \sqrt{x-2}+2&=x\\ \sqrt{x-2}&=x-2\\ x-2&=(x-2)^2\\ (x-2)^2-(x-2)&=0\\ (x-2)(x-2-1)&=0\\ (x-2)(x-3)&=0\\ x=2 \qquad\text{or}&\qquad x=3 \end{aligned} Because squaring is not an equivalent transformation in general, the solutions of the resulting equations must be checked. When $$x=2$$, the left side of the original equation is $$\sqrt{2-2}+2=0+2=2$$. When $$x=3$$, the left side is $$\sqrt{3-2}+2=1+2=3$$. So both $$x=2$$ and $$x=3$$ are solutions of the function $$\sqrt{x-1}+2=x$$.

## 8.5 Practice

Problem 8.1 Solve each radical equation.

1. $$\sqrt{3x+1}=4$$
2. $$\sqrt{2x-1}-5=0$$

Problem 8.2 Solve each radical equation.

1. $$\sqrt{5x+1}=x+1$$
2. $$x=\sqrt{3x+7}-3$$

Problem 8.3 Solve each radical equation.

1. $$\sqrt{6x+7}-x=2$$
2. $$\sqrt{x+2}+\sqrt{x-1}=3$$

Problem 8.4 Solve each radical equation.

1. $$\sqrt{x+5}-\sqrt{x-3}=2$$
2. $$3\sqrt[3]{3x-1}=6$$

Problem 8.5 Solve each radical equation.

1. $$(x+3)^{\frac12}=x+1$$
2. $$2(x-1)^{\frac12}-(x-1)^{-\frac12}=1$$

Problem 8.6 Solve each radical equation.

1. $$(x-1)^{\frac32}=8$$
2. $$(x+1)^{\frac23}=4$$