# 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 \(\sqrt[n]{X}=a\) 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. *

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

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

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

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. *

- Isolated one radical. \[ \sqrt{x-1}=\sqrt{x-6}+1\\ \]
- Square both sides to remove radical sign and then isolate the remaining radical. \[ \begin{aligned} x-1&=(x-6)+2\sqrt{x-6}+1\\ x-1&=x-5+2\sqrt{x-6}\\ 4&=2\sqrt{x-6}\\ 2&=\sqrt{x-6}. \end{aligned} \]
- 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. *

- Isolated the radical. \[ \sqrt[3]{x-4}=-3 \]
- 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.

- \(\sqrt{3x+1}=4\)
- \(\sqrt{2x-1}-5=0\)

**Problem 8.2 **
Solve each radical equation.

- \(\sqrt{5x+1}=x+1\)
- \(x=\sqrt{3x+7}-3\)

**Problem 8.3 **
Solve each radical equation.

- \(\sqrt{6x+7}-x=2\)
- \(\sqrt{x+2}+\sqrt{x-1}=3\)

**Problem 8.4 **
Solve each radical equation.

- \(\sqrt{x+5}-\sqrt{x-3}=2\)
- \(3\sqrt[3]{3x-1}=6\)

**Problem 8.5 **
Solve each radical equation.

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

**Problem 8.6 **
Solve each radical equation.

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