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, 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{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