Topic 8 Radical Equations

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 $\sqrt[n]{nX}=a$ 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, you’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. $$\begin{array}{cc}{x}^2-2x+1& =x+1\\ x^2-3x& =0\\ x(x-3)& =0\end{array}$$ $$\begin{array}{ccc}x=0& \text{or}& x-3=0\\ x=0& \text{or}& x=3\end{array}$$

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. $$\begin{array}{cc}x-1& =(x-6)+2\sqrt{x-6}+1\\ x-1& =x-5+2\sqrt{x-6}\\ 4& =2\sqrt{x-6}\\ 2& =\sqrt{x-6}.\end{array}$$
3. Square both sides to remove the radical sign and then solve. $$\begin{array}{cc}\sqrt{x-6}& =2\\ x-6& =4\\ x& =10.\end{array}$$ 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 $-2\sqrt[3]{3x-4}=6.$

Solution.

1. Isolated the radical. $$\sqrt[3]{3x-4}=-3$$
2. Cube both sides to eliminate the cube root and then solve the resulting equation. $$\begin{array}{cc}x-4& =(-3{)}^3\\ x-4& =-27\\ x& =-23\end{array}$$ 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 ${\left(X^{\frac{m}{n}}\right)}^{\frac{n}{m}}$. When $m$ is even and $n$ is odd, ${\left(X^{\frac{m}{n}}\right)}^{\frac{n}{m}}=|X|$. Otherwise, ${\left(X^{\frac{m}{n}}\right)}^{\frac{n}{m}}=X$.

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

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. $$\begin{array}{cc}(x+2{)}^{\frac{1}{2}}-(x-3{)}^{\frac{1}{2}}=& 1\\ (x+2{)}^{\frac{1}{2}}=& (x-3{)}^{\frac{1}{2}}+1\\ x+2=& {\left(\right(x-3{)}^{\frac{1}{2}}+1)}^2\\ x+2=& (x-3)+2(x-3{)}^{\frac{1}{2}}+1\\ 2(x-3{)}^{\frac{1}{2}}=& 4\\ (x-3{)}^{\frac{1}{2}}=& 2\\ x-3=& 4\\ x=& 7\end{array}$$ Check: $$(7+2{)}^{\frac{1}{2}}-(7-3{)}^{\frac{1}{2}}=\sqrt{9}-\sqrt{4}=3-2=1.$$

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. $$\begin{array}{cc}(x-1{)}^{\frac{2}{3}}=& 4\\ {\left(\right(x-1{)}^{\frac{2}{3}})}^{\frac{3}{2}}=& 4^{\frac{3}{2}}\\ |x-1|=& 8\\ x-1=8\text{or}& x-1=-8\\ x=9\text{or}& x=-7\end{array}$$ Check: $$(9-1{)}^{\frac{2}{3}}=8^{\frac{2}{3}}=(8^{\frac{1}{3}}{)}^2=2^2=4;$$ $$(-7-1{)}^{\frac{2}{3}}=(-8{)}^{\frac{2}{3}}=\left(\right(-8{)}^{\frac{1}{3}}{)}^2=(-2{)}^2=4.$$

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): $$\begin{array}{cc}\sqrt{x-2}+2& =x\\ (\sqrt{x-2}{)}^2+2^2& =x^2\\ x-2+4& =x^2\\ x+2& =x^2\\ x^2-x-2& =0\\ (x-2)(x+1)& =0\\ x-2=0\text{or}& x+1=0\\ x=2\text{or}& x=-1\end{array}$$ 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. $$\begin{array}{cc}\sqrt{x-2}+2& =x\\ \sqrt{x-2}& =x-2\\ x-2& =(x-2{)}^2\\ (x-2{)}^2-(x-2)& =0\\ (x-2)(x-2-1)& =0\\ (x-2)(x-3)& =0\\ x=2\text{or}& x=3\end{array}$$ 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. $3\sqrt[3]{33x-1}=6$

Problem 8.5 Solve each radical equation.

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

Problem 8.6 Solve each radical equation.

1. $(x-1{)}^{\frac{3}{2}}=8$
2. $(x+1{)}^{\frac{2}{3}}=4$