Solving Radical Equations

Equations with variable terms inside a radical are called radical equations. Like all equations, these can be solved graphically and algebraically. Sometimes, when solving algebraically, extraneous solutions — or solutions that do not satisfy the equation — arise.

Method

Solving Radical Equations Graphically

A radical function is a function that contains a radical expression, such as $y = 3 2 + x + 5 .$ If the dependent variable of the function is exchanged for a constant, say $C,$ the result is radical equation, $C = 3 2 + x + 5,$

which can be solved graphically. This is done by first graphing the function

y = 3 2 + x + 5,

then finding the

x

-coordinate of the point on the graph that has the

y

-coordinate

C .

Then, the

x

-coordinate is the solution to the equation.

Solve the radical equation graphically

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Exercise

Solve the equation graphically. $3 2 x + 2 = x - 1$

Show Solution

Solution

When solving an equation graphically it is necessary to have all variables on one side. Therefore, we'll first rearrange the equation. $3 2 x + 2 = x - 1 \Leftrightarrow 3 2 x + 2 - x = - 1$ The expression on the left-hand side can now be seen as the function, $f (x) = 3 2 x + 2 - x .$ We'll graph the function.

Next, we can identify the points that have the $y$ -coordinate $- 1,$ then find the corresponding $x$ -coordinates.

Since the point

(3, - 1)

is a point on

f,

x = 3

is a solution to the equation. We can verify this by testing it in the equation.

3 2 x + 2 = x - 1

Substitute

x = 3

3 2 \cdot 3 + 2 = ? 3 - 1

MultiplyMultiply

3 6 + 2 = ? 3 - 1

AddSubTermsAdd and subtract terms

38 = ? 2

CalcRootCalculate root

2 = 2

Thus,

x = 3

is a solution to the equation.

Method

Solving a Radical Equation Algebraically

Radical equations can be solved algebraically using inverse operations. Specifically, to undo the radical, both sides of the equation can be raised to the same power as the index of the radical. For example,

(3 x + 2)^{3} \Leftrightarrow x + 2 .

Because some radicals can only take certain

x

-values, this process can produce extraneous solutions, or solutions that do not actually satisfy the equation. Therefore, each solution must be verified in the original equation.
Consider the following equation as an example.

2 = x + 2 x - 1

1

Isolate the radical on one side

When solving a radical equation, it is necessary to isolate the radical on one side before raising the equation to an exponent. Using inverse operations, the equation becomes $2 = x + 2 x - 1 \Leftrightarrow 2 - x = 2 x - 1 .$

2

Eliminate the radical

Now, the radical can be undone or eliminated by raising it to the same power as the index of the radical. Here, the radical is a square root, so the index is

2 .

Thus, both sides of the equation are raised to the power of

2 .

2 - x = 2 x - 1

RaiseEqn

LHS^{2} = RHS^{2}

(2 - x)^{2} = (2 x - 1)^{2}

PowSqrt

(a)^{2} = a

(2 - x)^{2} = 2 x - 1

3

Solve the equation

When the radical has been eliminated the resulting equation can be solved.

(2 - x)^{2} = 2 x - 1

ExpandNegPerfectSquare

(a - b)^{2} = a^{2} - 2 a b + b^{2}

2^{2} - 2 \cdot 2 x + x^{2} = 2 x - 1

CalcPowProdCalculate power and product

4 - 4 x + x^{2} = 2 x - 1

Now, the equation can be solved for the variable. Notice here that the highest exponent on

x

is

2 .

That means this is a quadratic equation, and must be solved as such. It'll be set equal to

0

before using the Quadratic Formula.

4 - 4 x + x^{2} = 2 x - 1

SubEqn

LHS - 2 x = RHS - 2 x

4 - 6 x + x^{2} = - 1

AddEqn

LHS + 1 = RHS + 1

5 - 6 x + x^{2} = 0

RearrangeEqnRearrange equation

x^{2} - 6 x + 5 = 0

UseQuadFormUse the Quadratic Formula:

a = 1, b = - 6, c = 5

x = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

NegNeg

- (- a) = a

x = \frac{6 \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

CalcPowProdCalculate power and product

x = \frac{6 \pm 3 6 - 2 0}{2}

SubTermSubtract term

x = \frac{6 \pm 1 6}{2}

CalcRootCalculate root

x = \frac{6 \pm 4}{2}

StateSolState solutions

x = \frac{6 + 4}{2} x = \frac{6 - 4}{2}

AddSubTermsAdd and subtract terms

x = \frac{1 0}{2} x = \frac{2}{2}

CalcQuotCalculate quotient

x = 5 x = 1

The

x

-values for the solutions to the equation are

x = 5

and

x = 1 .

4

Check for extraneous solutions

The solutions found in Step

3

might be extraneous solutions. Thus, they must be verified in the original equation. First the solution

x = 5

is tested.

2 = x + 2 x - 1

Substitute

x = 5

2 = ? 5 + 2 \cdot 5 - 1

MultiplyMultiply

2 = ? 5 + 10 - 1

SubTermsSubtract terms

2 = ? 5 + 9

CalcRootCalculate root

2 = ? 5 + 3

AddTermsAdd terms

2 \neq = 8

Since we arrived at a contradiction,

x = 5

does not satisfy the radical equation. Thus, it is an extraneous solution.

x = 1

can be checked in the same way.

2222 = ? 1 + 2 \cdot 1 - 1 = ? 1 + 2 - 1 = ? 1 + 1 = 2

Since $x = 1$ makes a true statement, it is a solution to the radical equation. With regard to discussing solutions, it can be said that $2 = x + 2 x - 1$ has one solution and one extraneous solution.

Solve the radical equation

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Exercise

The Lorenz factor, $γ,$ is used for calculations within the Theory of Relativity. It is defined as $γ = \frac{1}{1 - β ^{2}}, β > 0 .$ Find $β,$ when $γ = 1.25 .$

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Solution

Here, the Greek letters gamma,

γ,

and beta,

β,

are used to denote the variables. We can solve for

β

by substituting

γ = 1.25

into the equation and using inverse operations to isolate

β .

γ = \frac{1}{1 - β ^{2}}

Substitute

γ = 1.25

1.25 = \frac{1}{1 - β ^{2}}

MultEqn

LHS \cdot 1 - β^{2} = RHS \cdot 1 - β^{2}

1.25 \cdot 1 - β^{2} = 1

DivEqn

LHS / 1.25 = RHS / 1.25

1 - β^{2} = \frac{1}{1 . 2 5}

UseCalcUse a calculator

1 - β^{2} = 0.8

To eliminate the radical, both sides of the equation can be squared.

1 - β^{2} = 0.8

RaiseEqn

LHS^{2} = RHS^{2}

(1 - β^{2})^{2} = 0 . 8^{2}

PowSqrt

(a)^{2} = a

1 - β^{2} = 0 . 8^{2}

CalcPowCalculate power

1 - β^{2} = 0.64

SubEqn

LHS - 1 = RHS - 1

- β^{2} = - 0.36

DivEqn

LHS / (- 1) = RHS / (- 1)

β^{2} = 0.36

SqrtEqn

LHS = RHS

β = \pm 0.36

β > 0

β = 0.36

CalcRootCalculate root

β = 0.6

Lastly, we can determine if

β = 0.6

is an extraneous solution.

1.25 = \frac{1}{1 - β ^{2}}

Substitute

β = 0.6

1.25 = ? \frac{1}{1 - 0 . 6 ^{2}}

CalcPowCalculate power

1.25 = ? \frac{1}{1 - 0 . 3 6}

SubTermSubtract term

1.25 = ? \frac{1}{0 . 6 4}

CalcRootCalculate root

1.25 = ? \frac{1}{0 . 8}

UseCalcUse a calculator

1.25 = 1.25

As

β = 0.6

made a true statement, it is a solution to the equation.

Solving Radical Equations