College Algebra
Answer/Discussion to Practice Problems
Tutorial 17: Quadratic Equations
WTAMU > Virtual Math Lab > College Algebra > Tutorial 17: Quadratic Equations
Answer/Discussion
to 1a

This quadratic equation is already simplified. 
Step 2: Write
in standard form, ,
if needed. 
This quadratic equation is already in standard form. 

*Use ZeroProduct Principle
*Solve the first linear equation
*Solve the second linear equation

There are two solutions to this quadratic equation: x =
4 and x = 5. 
Answer/Discussion
to 1b


*Use Dist. Prop. to clear the ( ) 
Step 2: Write
in standard form, ,
if needed. 

*Inverse of add. 9 is sub. 9
*Quad. eq. in standard form


*Use ZeroProduct Principle
*Solve the first linear equation
*Solve the second linear equation

There are two solutions to this quadratic equation: x =
3/7 and x = 3/2. 
Answer/Discussion
to 2a

Note how this quadratic equation is not in the form to begin with. The 3 is NOT part of the expression being squared
on the left side of the equation. We can easily write it in the form by dividing both sides by 3. 

*Not in the form
*Inv. of mult. by 3 is div. by 3
*Written in the form
*Apply the sq. root method
*There are 2 solutions 
Step 3: Solve for the
linear equation(s) set up in step 2. 

*Sq. root of 25 = 5
*Neg. sq. root of 25 = 5

There are two solutions to this quadratic equation: x =
5 and x = 5. 
Answer/Discussion
to 2b


*Written in the form
*Apply the sq. root method
*There are 2 solutions 
Step 3: Solve for the
linear equation(s) set up in step 2. 

*Sq. root of 12 = 2 sq. root of 3
*Solve for x
*Neg. sq. root of 12 = 2 sq. root of 3
*Solve for x

There are two solutions to this quadratic equation: x =
and x = . 
Answer/Discussion
to 3a

The coefficient of the term is already 1. 
Note how the and x terms are not isolated to begin with.
We can easily fix that by moving the constant to the other side of the
equation. 

*Inverse of add. 13 is sub. 13
* and x terms are now isolated


*b is the coefficient
of the x term
*Complete the square by taking 1/2 of b and squaring it


*Add constant found above to BOTH sides of
the eq.
*This creates a PST on the left side of eq. 

*Written in the form
*Apply the sq. root method
*There are 2 solutions

There are two solutions to this quadratic equation: x =
1 and x = 13. 
Answer/Discussion
to 3b

Note how the coefficient on the term is not 1 to begin with. We can easily fix that by dividing both
sides by that coefficient, which in this case is 5 . 

*Divide both sides by 5
*Coefficient of term is now 1

The and x terms
are already isolated. 

*b is the coefficient of the x term
*Complete the square by taking 1/2 of b and squaring it


*Add constant found above to BOTH sides of
the eq.
*This creates a PST on the left side of eq.


*Written in the form
*Apply the sq. root method
*There are 2 solutions

There are two solutions to this quadratic equation: x =
and x = . 
Answer/Discussion
to 4a

This quadratic equation is already simplified. 
Step 2: Write
in standard form, ,
if needed. 
This quadratic equation is already in standard form. 
a, the number in front of x squared, is 1.
b, the number in front of x,
is 10.
c, the constant, is 25.
Make sure that you keep the sign that is in front of each of these numbers.
Next we will plug it into the quadratic formula. Note that
we are only plugging in numbers, we don't also plug in the variable. 
AND
Step 5: Simplify if possible. 

*Quadratic formula
*Plug in values found above for a, b,
and c
*Simplify

Answer/Discussion
to 4b

This quadratic equation is already simplified. 
Step 2: Write
in standard form if needed. 
This quadratic equation is already in standard form. 
a, the number in front of x squared, is 5.
b, the number in front of x,
is 2.
c, the constant, is 10.
Make sure that you keep the sign that is in front of each of these numbers.
Next we will plug it into the quadratic formula. Note that
we are only plugging in numbers, we don't also plug in the variable. 
AND
Step 5: Simplify if possible. 
Answer/Discussion
to 4c

This quadratic equation is already simplified. 
Step 2: Write
in standard form if needed. 

*Inverse of add. 7x and 20 is sub. 7x and 20
*Quad. eq. in standard form 
a, the number in front of x squared, is 3.
b, the number in front of x,
is 7.
c, the constant, is 20.
Make sure that you keep the sign that is in front of each of these numbers.
Next we will plug it into the quadratic formula. Note that
we are only plugging in numbers, we don't also plug in the variable. 
AND
Step 5: Simplify if possible. 

*Quadratic formula
*Plug in values found above for a, b,
and c
*Simplify

Answer/Discussion
to 5a

This quadratic equation is already simplified. 
Step 2: Write
in standard form, ,
if needed. 
This quadratic equation is already in standard form. 
a, the number in front of x squared, is 1.
b, the number in front of x,
is 12.
c, the constant, is 36.
Make sure that you keep the sign that is in front of each of these numbers. 
Step 4: Plug the
values found in step 3 into the discriminant, ,
AND
Step 5: Simplify if possible. 

*Discriminant formula
*Plug in values found above for a, b,
and c
*Discriminant 
Since the discriminant is zero, that means there is only one real
number solution. 
Answer/Discussion
to 5b

This quadratic equation is already simplified. 
Step 2: Write
in standard form, ,
if needed. 

*Inverse of add. x is sub. x
*Quad. eq. in standard form 
a, the number in front of x squared, is 5.
b, the number in front of x,
is 1.
c, the constant, is 0.
Note that since the constant is missing it is understood to be 0.
Make sure that you keep the sign that is in front of each of these numbers. 
Step 4: Plug the
values found in step 3 into the discriminant, ,
AND
Step 5: Simplify if possible. 

*Discriminant formula
*Plug in values found above for a, b,
and c
*Discriminant 
Since the discriminant is a positive number, that means there are
two distinct real number solutions. 
Answer/Discussion
to 5c

This quadratic equation is already simplified. 
Step 2: Write
in standard form, ,
if needed. 

*Inverse of sub. x and
12 is add. x and 12
*Quad. eq. in standard form 
a, the number in front of x squared, is 2.
b, the number in front of x,
is 1.
c, the constant, is 12.
Make sure that you keep the sign that is in front of each of these numbers. 
Step 4: Plug the
values found in step 3 into the discriminant, ,
AND
Step 5: Simplify if possible. 

*Discriminant formula
*Plug in values found above for a, b,
and c
*Discriminant 
Since the discriminant is a negative number, that means there are
two distinct complex imaginary solutions. 
Last revised on Dec. 16, 2009 by Kim Seward.
All contents copyright (C) 2002  2010, WTAMU and Kim Seward.
All rights reserved.

