Download Lesson 6_5

Lesson 6.5 Summary Solving Square Roots and Other Radical
Expressions
Objective: To solve square root and other radical equations.
MA.912.A.6.5, MA.912.A.6.4.
Introduction:
To solve a radical equation, isolate the radical on one side of the
equation. Then raise each side to the power suggested by the index.
This can introduce extraneous solutions, or solutions that do not make
the original equation a true statement.
A radical equation is an equation that has a variable in a radicand or a
variable with a rational exponent. If the radical is 2, the equation is a
square root equation. In this lesson, assume that all radicals and
expressions with rational exponents represent real numbers.
Notice that when you take the square root of each side of an equation,
for example, you do need to introduce a ± sign. This is because the root
of any even index could be either positive or negative, so in general,
you introduce a ± sign when the index is even. However, when you
raise both sides of an equation to either an even or odd power, you do
not have to include a ± sign.
𝑚
𝑛
To solve equations of the form 𝑥 = 𝑘, raise each side of the equation
𝑛
𝑚
to the power , the reciprocal of . This is done to eliminate the
𝑚
𝑛
𝑛
𝑚 𝑚
𝑛
rational exponent. However, if either m or n is even, then (𝑥 ) =
|𝑥|. This is because if you raise the radicand to an even power, it will
always result in a positive number, whether the value in the radicand is
positive or negative, and that positive number can have a negative and
a positive root depending on whether the index is even or not. But
since we are only interested in the principal, or positive, root, the value
of the variable included in the expression for the root cannot be
negative. On the other hand, if the index is even, the value in the
radicand can have a positive and a negative root. But again, since we
are only interested in the principal, or positive, root, the value of the
variable included in the expression for the root cannot be negative. It is
for these reasons, that the variable is represented within the absolute
value sign in both cases.
When solving radical equations with a rational exponent, isolate the
exponential expression then use the inverse of the power to simplify
and solve the equation.
Isolating the radical or exponential expression is important because if
you raise each side of the original equation to a power in order to
remove the radical or the rational exponent you will end up with a
more complicated equation, not a simpler one.
Notice that when you have a radical equation whose radical expression
can be converted to an exponential expression with a numerator of
other than 1, it is best to convert the radical expression to an
exponential expression and then raise it to the reciprocal of the rational
exponent in order to remove the rational exponent.
When you raise each side of an equation to a power, it is possible to
introduce extraneous solutions. Therefore, it is important that you
check all solutions in the original equation. An extraneous solution will
give a false statement.
To square a binomial, use the formula: (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 or
(𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 .
If an equation contains two radical expressions (or two terms with
rational exponents), isolate one of the radicals (or one of the terms),
then eliminate it (or its rational exponent). Isolate the more
complicated radical expression first. In the resulting equation,
simplify the expressions before you eliminate the second radical.
In lesson 6.8 you will study the graphs of square root functions. These
graphs can help you find solutions and identify extraneous solutions.
To graph, for example, the equation √𝑥 + 7 − 5 = 𝑥, let Y1=
√ (𝑥 + 7) − 7 𝑎𝑛𝑑 𝑌2 = 𝑥. The solutions to the equation are the x
values where the graphs of the two equations intercept. The
extraneous solutions are those x values that could be considered as
possible solutions but for which the graphs do not intercept.
Demonstrated: Got It 1 pg. 391:
What is the solution of √4𝑥 + 1 − 5 = 0?
Classwork: ex. 13 pg. 395:
Solve: √2𝑥 − 1 = 3
Demonstrated: Got It 2 pg. 392:
2
What are the solution(s) of 2(𝑥 + 3)3 = 8?
Classwork: ex. 19 pg. 395:
2
3
Solve: (𝑥 + 2) = 9
Demonstrated: Got It 4a pg. 393:
What is the solution of √5𝑥 − 1 + 3 = 𝑥?
Classwork: ex. 27 pg. 395:
1
2
Solve: (5 − 𝑥 ) = 𝑥 + 1
Demonstrated: Got It 5 pg. 394:
What is the solution of √5𝑥 + 4 − √𝑥 = 4?
Classwork: ex. 38 pg. 395:
Solve: √3𝑥 + 2 − √2𝑥 + 7 = 0
Homework: pgs. 395-396 ex: 14, 22, 24, 26, 37, 46, 53, 56, 64, 67.