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4.04 Radical equations and inequalities

Lesson

Concept summary

Radical equation

An equation containing at least one radical expression

The process of solving radical equations is essentially the same as solving a linear equation, that is, by using inverse operations. The inverse of a square root operation is squaring, and the inverse of an nth root, is to take the nth power.

x
y

It is important to remember that the parent function f\left(x\right)=\sqrt{x} has a domain of x\geq0, and a range of y\geq0. When we are solving equations involving square roots, and other even roots, we need to remember that on every step of our solution, any radicands must be non-negative and any radicals must also be non-negative.

It is important to note, however, that the solution to the equation itself can be negative.

Extraneous solutions

When we square both sides we may get a solution that results in a negative radicand, or a solution which will result in a false statement if substituted into the original equation.

When solving questions with real life applications, we also need to ensure we have viable solutions, which make sense within the context of the question. A non-viable solution does not make sense within the context of the question, such as a negative value when we are solving for the length of a physical object.

Radical inequality

An inequality containing at least one radical expression

The process of solving radical inequalities is similar to solving a radical equation using inverse operations and properties to isolate the variable. However, there are two things to remember:

  1. As per the multiplication and division property of inequality, when multiplying or dividing an inequality by a negative value the inequality symbol is reversed
    \displaystyle -2\sqrt{x}\displaystyle <\displaystyle -6
    \displaystyle \sqrt{x}\displaystyle >\displaystyle \dfrac{-6}{-2}Division property of inequality: notice the sign is reversed
  2. A radical equation with an even index and a negative radicand has no real roots, so we must also consider the additional restriction that the radicand must be positive for even indexes. \text{If }\sqrt{2x+3}<7 \text{, then we have to consider that }2x+3 \geq 0

We can solve radical inequalities algebraically, ensuring to test the solution set found is valid, or we can solve with technology by graphing or creating a table of values.

Worked examples

Example 1

Solve the following equation:

\sqrt{y}+5=9

Approach

To solve for y, we first want to isolate \sqrt{y}, we can then square both sides of the equation.

Solution

\displaystyle \sqrt{y}+5 \displaystyle =\displaystyle 9State the equation
\displaystyle \sqrt{y} \displaystyle =\displaystyle 4Subtraction property of equality
\displaystyle \left(\sqrt{y}\right)^2\displaystyle =\displaystyle 4^2Square both sides
\displaystyle y\displaystyle =\displaystyle 16Evaluate the squares

Reflection

We can see that if we substitute y=16 into the original equation we obtain \sqrt{16}+5=9 which is a true statement, and as \sqrt{16} is positive we have a valid solution.

Example 2

Solve each equation for x. Identify any extraneous solutions.

a

3x=1+2\sqrt{x}

Approach

To solve for x, we first want to isolate 2\sqrt{x}. We can then square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

Solution

\displaystyle 3x\displaystyle =\displaystyle 1+2\sqrt{x}State the equation
\displaystyle 3x-1\displaystyle =\displaystyle 2\sqrt{x}Subtraction property of equality
\displaystyle \left(3x-1\right)^2\displaystyle =\displaystyle \left(2\sqrt{x}\right)^2Square both sides
\displaystyle 9x^2-6x+1\displaystyle =\displaystyle 4xEvaluate the squares
\displaystyle 9x^2-10x+1\displaystyle =\displaystyle 0Subtraction property of equality
\displaystyle \left(9x-1\right)\left(x-1\right)\displaystyle =\displaystyle 0Factor the quadratic

Using the zero product rule we get two solutions x=\dfrac{1}{9}, x=1. We now want to test both solutions:

\displaystyle 3\left(1\right)\displaystyle =\displaystyle 1+2\sqrt{1}Substitute x=1
\displaystyle 3\displaystyle =\displaystyle 3Evaluate both sides

We can see that x=1 satisfies the original equation and is a valid solution.

Now, let's test x=\dfrac{1}{9}

\displaystyle 3\left(\dfrac{1}{9}\right)\displaystyle =\displaystyle 1+2\sqrt{\frac{1}{9}}Substitute x=\dfrac{1}{9}
\displaystyle \frac{1}{3}\displaystyle =\displaystyle \frac{5}{3}Evaluate both sides

We can see that x=\dfrac{1}{9} leads to a false statement and is therefore an extraneous solution.

b

\sqrt{x+17}=x+5

Approach

To solve for x, we first want to square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

Solution

\displaystyle \sqrt{x+17}\displaystyle =\displaystyle x+5State the equation
\displaystyle \left(\sqrt{x+17}\right)^2\displaystyle =\displaystyle \left(x+5\right)^2Square both sides
\displaystyle x+17\displaystyle =\displaystyle x^2+10x+25Evaluate the squares
\displaystyle 0\displaystyle =\displaystyle x^2+9x+8Subtraction property of equality
\displaystyle 0\displaystyle =\displaystyle \left(x+1\right)\left(x+8\right)Factor the quadratic

Using the zero product rule we get two solutions x=-8, x=-1. We now want to test both solutions:

\displaystyle \sqrt{-8+17}\displaystyle =\displaystyle -8+5Substitute x=-8
\displaystyle 3\displaystyle =\displaystyle -3Evaluate both sides

We can see that x=-8 leads to a false statement and is therefore an extraneous solution.

Now, let's test x=-1.

\displaystyle \sqrt{-1+17}\displaystyle =\displaystyle -1+5Substitute x=-1
\displaystyle 4\displaystyle =\displaystyle 4Evaluate both sides

We can see that x=-1 satisfies the original equation and is a valid solution.

Reflection

It is important to note that the fact that our solutions were negative, x=-8, x=-1, does not necessarily make them extraneous. We can see after substitution that only x=- 8 is an extraneous solution, and x=-1 is valid.

Example 3

Solve each inequality for x. Represent the solution set on a number line.

a

\sqrt{3x+12} \lt 6

Approach

  1. Consider the inequality \sqrt{3x+12} \lt 6 and isolate x using inverse operations.

  2. Consider the additional condition that the radicand must be positive by solving the inequality 3x+12 \geq 0 for x.

  3. Find the solution of the intersection of the two intervals found and represent the solution graphically.

  4. Check the solution set includes all viable solutions.

Solution

Step 1. Consider the inequality \sqrt{3x+12} \lt 6 and isolate x:

\displaystyle \sqrt{3x+12}\displaystyle <\displaystyle 6State the inequality
\displaystyle \left(\sqrt{3x+12}\right)^2\displaystyle <\displaystyle \left(6\right)^2Square both sides
\displaystyle 3x+12\displaystyle <\displaystyle 36Evaluate the squares
\displaystyle 3x\displaystyle <\displaystyle 24Subtraction property of inequality
\displaystyle x\displaystyle <\displaystyle 8Division property of inequality

Step 2. Consider the inequality 3x+12 \geq 0 and isolate x:

\displaystyle 3x+12\displaystyle \geq\displaystyle 0State the inequality
\displaystyle 3x\displaystyle \geq\displaystyle -12Subtraction property of inequality
\displaystyle x\displaystyle \geq\displaystyle -4Division property of inequality

Step 3. The solution set will be all the numbers that make both inequalities true, that is, both x \geq -4 and x \lt 8. This can be written as a single inequality as -4 \leq x \lt 8 and represented graphically on a number line as follows:

-10-9-8-7-6-5-4-3-2-1012345678910

Step 4. Notice the two points x=-4 and x=8 divide the number line into three sections. Numbers less than -4 will make the radicand negative, so cannot be part of the solution set. We can double check the correct interval is highlighted by checking a point inside the other two sections to determine if it is part of the solution set. For example, we can test x=0 and x=10.

\displaystyle \sqrt{3\left(0\right)+12}\displaystyle <\displaystyle 6Substitute x=0
\displaystyle \sqrt{12}\displaystyle <\displaystyle 6Evaluate both sides

We can see that x=0 leads to a true statement and is therefore a valid solution along with other points in the interval -4 \leq x\lt 8.

\displaystyle \sqrt{3\left(10\right)+12}\displaystyle <\displaystyle 6Substitute x=10
\displaystyle \sqrt{42}\displaystyle <\displaystyle 6Evaluate both sides

We can see that x=10 leads to a false statement and is therefore not a valid solution along with other points in the interval x\geq 8.

Reflection

When creating the number line note the signs of the inequality and use an unfilled circle when an endpoint is not included and a filled circle when an endpoint is included.

As squaring both sides of an equation or inequality may introduce extraneous solutions it is important to check we have the correct solution set.

b

\sqrt{3x-6}+1 \geq 2x-6, with technology

Approach

To solve using technology graph f\left(x\right)=\sqrt{3x-6}+1 and g\left(x\right)=2x-6 and determine when the function f\left(x\right) is equal or above the function g\left(x\right).

Solution

A screenshot of the GeoGebra graphing tool showing graphs of the functions f of x and g of x, together with their point of intersection (5, 4). Speak to your teacher for more details.

Enter the left side of the inequality as f\left(x\right) and the right side as g\left(x\right). Then use the intersection tool to find when the two functions are equal.

We can see f\left(x\right) has a domain of x\geq 2 and lies above g\left(x\right) on the interval 2 \leq x \leq 5.

This can be represented graphically on a number line as follows:

-2-1012345678

Reflection

Technology can assist to solve multi-step inequalities and to confirm the solution set when solving algebraically.

We could also have solved the above inequality algebraically as follows:

Consider the inequality \sqrt{3x-6}+1 \geq 2x-6 and isolate x:

\displaystyle \sqrt{3x-6}+1\displaystyle \geq\displaystyle 2x-6State the inequality
\displaystyle \sqrt{3x-6}\displaystyle \geq\displaystyle 2x-7Subtraction property of inequality
\displaystyle \left(\sqrt{3x-6}\right)^2\displaystyle \geq\displaystyle \left(2x-7\right)^2Square both sides
\displaystyle 3x-6\displaystyle \geq\displaystyle 2x^2-28x+49Evaluate the squares
\displaystyle 0\displaystyle \geq\displaystyle 4x^2-31x+55Subtraction property of inequality
\displaystyle 0\displaystyle \geq\displaystyle \left(4x-11\right)\left(x-5\right)Factor the quadratic

We can visualize the solution set by considering the graph of the corresponding quadratic equation:

1
2
3
4
5
x
-5
-4
-3
-2
-1
1
2
3
y

The graph shows the function {f\left(x\right)=\left(4x-11\right)\left(x-5\right)}, with roots of {x=\dfrac{11}{4}} and x=5.

The inequality is true when the quadratic expression is less than 0, so the solutions will be when the graph is below the x-axis.

The solution set will be \dfrac{11}{4} \leq x \leq 5. The inequality symbol tells us that the end points are included in the solution.

Then, consider the inequality representing the restriction on the radicand 3x-6 \geq 0 and isolate x:

\displaystyle 3x-6\displaystyle \geq\displaystyle 0State the inequality
\displaystyle 3x\displaystyle \geq\displaystyle 6Subtraction property of inequality
\displaystyle x\displaystyle \geq\displaystyle 2Division property of inequality

For inequalities when there is a variable not under a square root and we square both sides, we have to be particularly careful to check solutions. Check the interval by testing values in the original inequality.

01234567

Notice the three points x=2, x=\dfrac{11}{4}, and x=5 divide the number line into four sections. Numbers less than 2 will make the radicand negative, so cannot be part of the solution set. We can find the solution set by checking a point inside the other three sections to determine if it is part of the solution set. For example, we can test x=2.5, x=4 and x=6.

Checking these points in the inequality \sqrt{3x-6}+1 \geq 2x-6 creates a true statement for {x=2.5}, x=4 and a false statement for x=6. Thus, the solution set is: 2 \leq x \leq 5.

Outcomes

M2.N.Q.A.1

Use units as a way to understand real-world problems.*

M2.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

M2.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

M2.A.REI.A.1

Understand solving equations as a process of reasoning and explain the reasoning. Construct a viable argument to justify a solution method.

M2.A.REI.B.3

Solve radical equations in one variable, and identify extraneous solutions when they exist.

M2.MP1

Make sense of problems and persevere in solving them.

M2.MP2

Reason abstractly and quantitatively.

M2.MP3

Construct viable arguments and critique the reasoning of others.

M2.MP6

Attend to precision.

M2.MP7

Look for and make use of structure.

M2.MP8

Look for and express regularity in repeated reasoning.

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