Just as students learn to solve linear equations and quadratic equations, they will also need to learn

how to solve radical equations. To solve equations involving radicals, we must use what is known as the power rule, which states that if x, y and n are real numbers and x = y, then xn = yn. This is important because to solve radical equations we have to remove the radical. In order to remove the radical, we must raise the radical to the index of the radical. For example, if we have 3√(x+3), we raise 3√(x+3) to the 3rd power to remove the radical. To remove a radical involving a square root, we raise it to the 2nd power and so on.

Suppose we want to determine how long it takes for an object to hit the ground after being dropped from a certain height or we want to know the original cost of equipment that is currently valued at x dollars. These types of scenarios as well as problems in forestry, carpentry, supply and demand and others can be solved using radical equations.

**Example:** A baseball is hit straight up in the air and takes 4 seconds to hit the ground after reaching it’s highest point. What was the height (in feet) of the baseball at its highest point?

The distance an object will fall is given by the formula t = √(d/16) where t is the time and d is the distance in feet.

Substitute 4 for t and solve for d.

4 = √(d/16)

Square both sides of the equation to remove the radical.

42 = [√(d/16)]2

16 = d/16

d = 256 feet

**Example:** Suppose you purchased a car 3 years ago for $15,000. You wish to sell the car and want to know what the car is worth. The rate r at which the car has depreciated is given by the following formula r = 1 – 3√(v/c) where v is the current value and c is the original cost.

What is the value of the car if the rate of depreciation is 8%?

First we convert 8% to 0.08.

Substitute 0.08 for r and $15,000 for c and solve for v.

0.08 = 1 – 3√(v/15000)

Isolate the radical and cube both sides of the equation

0.08 – 1 = -3√(v/15000)

-0.92 = – 3√(v/15000)

(-0.92)3 = [- 3√(v/15000)]3

-0.7787 = -v/15000 ( -0.923 was solved using a calculator)

11,680.50 = v

Therefore the value of the car today is $11,680.50.

**Example:** Suppose that the number of hammers produced at a given price can be estimated by the formula S = √(8y), where S is the supply (in hundreds) and y is the cost (in dollars). The demand D for the lawn mowers can be estimated by the formula D = √(640 – 2y2). At what price will the supply equal the demand?

First we have to set S = D.

√(8y) = √(640 – 2y2).

Square both sides of the equation to remove the radicals.

[√(8y)]2 = [√(640 – 2y2)]2

8y = 640 – 2y2

0 = -2y2 – 8y + 640

0 = -2(y2 + 4y – 320)

0 = -2(y – 16)(y + 20)

y – 16 = 0 or y + 20 = 0

y = 16, -20

We can eliminate the solution, y = -20 because you can’t have a negative cost. Substitute 16 into the original equation to check the solution. It does check, therefore 16 is a solution and the cost where supply equals demand is $16.

This guide should help students understand how to use radical equations to solve problems in the real world.