College Algebra
Answer/Discussion to Practice Problems
Tutorial 20: Equations that are Quadratic in Form
WTAMU > Virtual Math Lab > College Algebra > Tutorial 20: Equations that are Quadratic in Form
Answer/Discussion
to 1a

Step 1: Write
in Standard Form, ,
if needed. 

*Inverse of add. 13 y squared is sub. 13 y squared
*Equation in standard form 
Below, I have the original equation rewritten in a way to show you
that it is quadratic in form. Note how when you square y squared you get y to the fourth, which is what
you have in the first term. 

*Rewriting original equation to show it is
quadratic in form
*Note that (y squared) squared = y to the fourth
*When in stand. form, let t = the expression following b. 
Next, we need to substitute t in for y squared in the original equation. 

*Original equation
*Substitute t in
for y squared 
Note how we ended up with a quadratic equation when we did our substitution.
From here, we need to solve the quadratic equation that we have created. 
You can use any method you want to solve the quadratic equation: factoring,
completing the square or quadratic formula.
I'm going to factor it to solve it. 

*Use ZeroProduct Principle
*Set 1st factor = 0 and solve
*Set 2nd factor = 0 and solve

In step 2 we used the substitution of t = y squared. 
Let's find the value(s) of y when t = 9/4: 

*First solution
*Second solution

Let's find the value(s) of y when t = 1: 

*First solution
*Second solution

Step 5: Check
your solutions. 
Let's double check to see if y = 3/2
is a solution to the original equation. 

*Plugging in 3/2 for y
*True statement

Since we got a true statement, y = 3/2
is a solution. 
Let's double check to see if y = 3/2
is a solution to the original equation. 

*Plugging in 3/2 for y
*True statement

Since we got a true statement, y = 3/2
is a solution. 
Let's double check to see if y = 1 is
a solution to the original equation. 

*Plugging in 1 for y
*True statement 
Since we got a true statement, y = 1
is a solution. 
Let's double check to see if y = 1
is a solution to the original equation. 

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

Step 1: Write
in Standard Form, ,
if needed. 
This equation is already in standard form. 
Below, I have the original equation rewritten in a way to show you
that it is quadratic in form. Note how when you square x to the 1/2 power you get x, which is what you
have in the first term. 

*Rewriting original equation to show it is
quadratic in form
*Note that (x to the 1/2 power) squared = x
*When in stand. form, let t = the expression following b. 
Next, we need to substitute t in for x to
the 1/2 power in the original equation. 

*Original equation
*Substitute t in
for x to the 1/2 power 
Note how we ended up with a quadratic equation when we did our substitution.
From here, we need to solve the quadratic equation that we have created. 
You can use any method you want to solve the quadratic equation: factoring,
completing the square or quadratic formula.
I'm going to factor it to solve it. 

*Use ZeroProduct Principle
*Set 1st factor = 0 and solve
*Set 2nd factor = 0 and solve

In step 2 we used the substitution of t = x to the 1/2 power. 
Let's find the value(s) of x when t = 2: 

*Inverse of taking it to the 1/3 power is
raising it to the 3rd power 
Let's find the value(s) of x when t = 1: 

*Inverse of taking it to the 1/2 power is
raising it to the 2nd power 
Step 5: Check
your solutions. 
Let's double check to see if x = 4 is
a solution to the original equation. 

*Plugging in 4 for x
*True statement 
Since we got a true statement, x = 4
is a solution. 
Let's double check to see if x = 1 is
a solution to the original equation. 

*Plugging in 1 for x
*True statement 
Since we got a true statement, x = 1
is a solution. 
There are two solutions to this equation: x = 4 and x = 1. 
Answer/Discussion
to 1c

Step 1: Write
in Standard Form, ,
if needed. 
This equation is already in standard form. 
Note how the original equation has the exact same expression in the
two ( )'s and that the first ( ) is squared and the 2nd (
) is to the one power. This equation is quadratic in form. 

*When in stand. form, let t = the expression following b. 
Next, we need to substitute t in for x minus 5 in the original equation. 

*Original equation
*Substitute t in
for x minus 5 
Note how we ended up with a quadratic equation when we did our substitution.
From here, we need to solve the quadratic equation that we have created. 
You can use any method you want to solve the quadratic equation: factoring,
completing the square or quadratic formula.
I'm going to factor it to solve it. 

*Use ZeroProduct Principle
*Set 1st factor = 0 and solve
*Set 2nd factor = 0 and solve

In step 2 we used the substitution of t = x  5. 
Let's find the value(s) of x when t = 7: 

*Plug in 7 for t
*Inverse of sub. 5 is add. 5 
Let's find the value(s) of x when t = 5: 

*Plug in 5 for t
*Inverse of sub. 5 is add. 5 
Step 5: Check
your solutions. 
Let's double check to see if x = 2
is a solution to the original equation. 

*Plugging in 2 for x
*True statement 
Since we got a true statement, x = 2
is a solution. 
Let's double check to see if x = 10
is a solution to the original equation. 

*Plugging in 10 for x
*True statement 
Since we got a true statement, x = 10
is a solution. 
There are two solutions to this equation: x = 2 and x = 10. 

