College Algebra
Answer/Discussion to Practice Problems
Tutorial 19: Radical Equations and Equations Involving Rational Exponents
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Answer/Discussion
to 1a


*Inverse of add. 4 is sub. 4
*Square root is by itself on one side of eq. 
If you square a square root, it will disappear. This is what
we want to do here so that we can get x out
from under the square root and continue to solve for it. 

*Inverse of taking the sq. root is squaring
it 
Step 3: If you
still have a radical left, repeat steps 1 and 2. 
No more radicals exist, so we do not have to repeat steps 1 and 2. 
Step 4: Solve
the remaining equation. 

*Inverse of sub. 10 is add. 10



Let's check to see if x = 26 is an extraneous
solution: 

*Plugging in 26 for x
*False statement 
Since we got a false statement, x =
26 is an extraneous solution.
There is no solution to this radical equation. 
Answer/Discussion
to 1b

The radical in this equation is already isolated. 
If you square a square root, it will disappear. This is what
we want to do here so that we can get y out
from under the square root and continue to solve for it. 

*Inverse of taking the sq. root is squaring
it
*Right side is a binomial
squared 
Be careful on this one. It
is VERY TEMPTING to square the right side term by term and get 4
+ ( x + 1). However, you need to square
it as a side as shown above. Recall that when you square a binomial
you get the first term squared minus twice the product of the two terms
plus the second term squared. If you need a review on squaring a
binomial, feel free to go to Tutorial
6: Polynomials. 
Step 3: If you
still have a radical left, repeat steps 1 and 2. 

*Inverse of add. x and 5 is sub. x and 5
*Square root is by itself on one side of eq.
*Inverse of taking the sq. root is squaring
it
*Left side is a binomial
squared

Be careful on this one. It
is VERY TEMPTING to square the left side term by term and get 4 x squared
plus 4. However, you need to square it as a side as shown above.
Recall that when you square a binomial you get the first term squared plus
twice the product of the two terms plus the second term squared.
If you need a review on squaring a binomial, feel free to go to Tutorial
6: Polynomials. 
Step 4: Solve
the remaining equation. 
In this example, the equation that resulted from squaring both sides
turned out to be a quadratic equation.
If you need a review on solving quadratic equations feel free to go
to Tutorial 17: Quadratic Equations. 
Let's check to see if x = 3 is
an extraneous solution: 

*Plugging in 3 for x
*True statement 
Since we got a true statement, x = 3
is a solution. 
Let's check to see if x = 1 is an extraneous
solution: 

*Plugging in 1 for x
*True statement 
Since we got a true statement, x = 1
is a solution.
There are two solutions to this radical equation: x = 3 and x = 1. 
Answer/Discussion
to 2a


*Inverse of sub. 9 is add. 9
*Inverse of mult. by 3 is div. by 3
*rat. exp. expression is by itself on one side
of eq. 
If you raise an expression that has a rational exponent to the reciprocal
of that rational exponent, the exponent will disappear. This
is what we want to do here so that we can get x out from under the rational exponent and continue to solve for it. 

*Inverse of taking it to the 5/2 power
is taking it to the 2/5 power

Step 3: Solve
the remaining equation. 
In this example, the equation that resulted from raising both sides
to the 2/5 power turned out to be a linear equation.
Also note that it is already solved for x.
So, we do not have to do anything on this step for this example. 
Let's check to see if is an extraneous solution: 

*Plugging in 3 to the 2/5 power for x
*True statement 
Since we got a true statement, is
a solution.
There is one solution to this rational exponent equation: . 
Answer/Discussion
to 2b

The base with the rational exponent is already isolated. 
If you raise an expression that has a rational exponent to the reciprocal
of that rational exponent, the exponent will disappear. This
is what we want to do here so that we can get x out from under the rational exponent and continue to solve for it. 

*Inverse of taking it to the 3/2 power is
taking it to the 2/3 power
*Use def.
of rat. exp

Step 3: Solve
the remaining equation. 
In this example, the equation that resulted from squaring both sides
turned out to be a quadratic equation.
If you need a review on solving quadratic equations, feel free to go
to Tutorial 17: Quadratic Equations. 
Let's check to see if x = 6 is an extraneous
solution: 

*Plugging in 6 for x
*True statement 
Since we got a true statement, x = 6
is a solution. 
Lets check to see if x = 3 is an extraneous
solution: 

*Plugging in 3 for x
*True statement 
Since we got a true statement, x = 3
is a solution.
There are two solutions to this rational exponent equation: x = 6 and x = 3. 

