Video Transcript
In this video, we will learn how to
solve quadratic and quadratic-like equations using factoring along with questions
which involve quadratics at some point in the factorization.
We begin by recalling the
definition of a quadratic equation. This is any equation that can be
expressed in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and
𝑐 are constants and 𝑎 is nonzero. We should already be familiar with
a number of methods for factoring quadratic expressions, including the following:
factoring by inspection, recognizing a quadratic as a perfect square, recognizing a
quadratic as the difference of two squares, factoring by grouping, and factoring by
completing the square. The focus of this video is the
application of these skills to solve quadratic equations.
Let’s suppose we have a quadratic
equation in its factored form 𝑝𝑥 plus 𝑞 multiplied by 𝑟𝑥 plus 𝑠 is equal to
zero, where 𝑝 and 𝑟 are not equal to zero. The key to solving such an equation
is to recognize that if the product of two or more factors is equal to zero, then at
least one of the individual factors must itself be equal to zero. This means that in order to find
all the solutions to the given equation, we set each factor equal to zero. This gives us two linear equations:
𝑝𝑥 plus 𝑞 equals zero and 𝑟𝑥 plus 𝑠 equals zero.
Solving the first equation by
subtracting 𝑞 from both sides and then dividing through by 𝑝, we have 𝑥 is equal
to negative 𝑞 over 𝑝. To solve the second equation, we
subtract 𝑠 from both sides and then divide through by 𝑟, giving us 𝑥 is equal to
negative 𝑠 over 𝑟. There are, therefore, two solutions
or roots to the given quadratic equation: 𝑥 is equal to negative 𝑞 over 𝑝 and 𝑥
is equal to negative 𝑠 over 𝑟. It is worth noting that in some
cases, the two solutions may coincide to give one repeated root, in which case we
only give this value once as the solution.
We will begin by considering an
example of solving a quadratic equation which is already in its factored form.
Solve the equation two 𝑥 minus
three multiplied by three 𝑥 plus four equals zero.
In this question, our quadratic
equation is already in a factored form. We are told that the product of two
linear expressions, two 𝑥 minus three and three 𝑥 plus four, is zero. And the only way this can happen is
if one or both of the factors individually is equal to zero. We have two possibilities: either
two 𝑥 minus three equals zero or three 𝑥 plus four equals zero.
We can solve the first equation by
firstly adding three to both sides. We can then divide through by two
such that 𝑥 is equal to three over two, or 1.5. We solve the second equation using
the same method. We subtract four from both sides
and then divide through by three. This gives us 𝑥 is equal to
negative four-thirds. The two solutions to the quadratic
equation two 𝑥 minus three multiplied by three 𝑥 plus four equals zero are 𝑥
equals three over two and 𝑥 equals negative four-thirds.
In our next example, we will
demonstrate how to solve a quadratic equation by first recognizing it as a perfect
square and, hence, writing the quadratic in its factored form.
Solve the equation 𝑥 squared minus
eight 𝑥 plus 16 equals zero by factoring.
There are many methods we could use
to factor the left-hand side of our quadratic equation. In this question, we will do so by
firstly recognizing that the first and third terms are perfect squares, since four
squared is equal to 16. We also notice that the middle term
negative eight 𝑥 is equal to the negative of twice the square root of the first
term multiplied by the square root of the third term. The quadratic expression is
therefore written in the form 𝑎 squared minus two 𝑎𝑏 plus 𝑏 squared. This means that the quadratic is a
perfect square and can be factored as 𝑎 minus 𝑏 multiplied by 𝑎 minus 𝑏, or more
simply as 𝑎 minus 𝑏 all squared.
We have already established that 𝑎
is equal to 𝑥 and 𝑏 is equal to four. And this means that 𝑥 squared
minus eight 𝑥 plus 16 can be rewritten as 𝑥 minus four all squared. We need to solve this expression
equal to zero. We have a repeated factor of 𝑥
minus four, and as such we have only one equation to solve: 𝑥 minus four equals
zero. Adding four to both sides of this
equation gives us our solution of 𝑥 is equal to four.
One way of checking our solution is
to substitute the value of 𝑥 equals four back in to the original equation. This gives us 16 minus 32 plus 16,
which is indeed equal to zero. The solution to the equation 𝑥
squared minus eight 𝑥 plus 16 equals zero is 𝑥 is equal to four.
Before moving on to our next
example, we will summarize the steps involved in solving a quadratic equation by
factoring. There is a three-step process we
can follow to solve a quadratic equation by factoring. Firstly, we find the factored form
of the quadratic equation using any available method. Our second step is to set each
factor equal to zero. Finally, we solve the resulting
equations. This process can also be applied to
higher-order equations whose factored forms involve quadratics. In our next example, we will
consider a cubic equation that can be factored into a linear term multiplied by a
quadratic and subsequently into the product of three linear expressions.
Find the solution set for two 𝑥
cubed is equal to 18𝑥.
The equation in this question is of
degree three and is, therefore, a cubic equation. We can solve the equation by
firstly subtracting 18𝑥 from both sides. This gives us two 𝑥 cubed minus
18𝑥 is equal to zero. Next, we observe that the two terms
on the left-hand side have a common factor of two 𝑥. The equation can therefore be
factored as two 𝑥 multiplied by 𝑥 squared minus nine is equal to zero.
We now have the product of a linear
term and a quadratic expression. The quadratic expression 𝑥 squared
minus nine is the difference of two squares, as it is written in the form 𝑎 squared
minus 𝑏 squared. We recall that this can be factored
into two linear expressions: 𝑎 plus 𝑏 and 𝑎 minus 𝑏. 𝑥 squared minus nine can therefore
be rewritten as 𝑥 plus three multiplied by 𝑥 minus three. And we have the equation two 𝑥
multiplied by 𝑥 plus three multiplied by 𝑥 minus three is equal to zero. We know that if the product of
three factors equals zero, at least one of the individual factors must be equal to
zero.
To find the solution set of the
equation, we need to solve the three equations two 𝑥 equals zero, 𝑥 plus three
equals zero, and 𝑥 minus three equals zero. This gives us three possible
solutions of 𝑥 equals zero, 𝑥 equals negative three, and 𝑥 equals three. The solution set for the equation
two 𝑥 cubed is equal to 18𝑥 is zero, negative three, and three. We could check each of these
solutions individually by substituting the values of 𝑥 back in to the original
equation.
The techniques we’ve encountered so
far can be applied to worded and practical problems. We will now consider one such
example.
Find the value of 𝑥 given that a
right triangle has a hypotenuse of length two 𝑥 and sides of lengths 𝑥 plus one
and 𝑥 plus three.
We will begin by sketching the
right triangle. We are told that the hypotenuse has
a length two 𝑥 and the other two sides have lengths 𝑥 plus one and 𝑥 plus
three. We can apply the Pythagorean
theorem to create a quadratic equation. This states that 𝑎 squared plus 𝑏
squared is equal to 𝑐 squared, where 𝑐 is the length of the hypotenuse and 𝑎 and
𝑏 are the lengths of the two shorter sides. Substituting in the given lengths,
we have 𝑥 plus one squared plus 𝑥 plus three squared is equal to two 𝑥
squared.
We can distribute the parentheses
on the left-hand side using the FOIL method. 𝑥 plus one all squared is equal to
𝑥 squared plus two 𝑥 plus one. And 𝑥 plus three all squared is
equal to 𝑥 squared plus six 𝑥 plus nine. The sum of these two expressions is
equal to four 𝑥 squared. By collecting like terms, the
left-hand side simplifies to two 𝑥 squared plus eight 𝑥 plus 10. We can then subtract all three of
these terms from both sides, leaving us with two 𝑥 squared minus eight 𝑥 minus 10
equals zero. Next, we can divide through by
two. We now have the quadratic equation
𝑥 squared minus four 𝑥 minus five equals zero, which we can solve by
factoring.
As the first term has a coefficient
of one, the first term in each of our parentheses will be 𝑥. We then need to find two integers
that have a product of negative five and a sum of negative four. Negative five multiplied by one is
negative five, and negative five plus one is equal to negative four. Our quadratic therefore factors to
𝑥 minus five multiplied by 𝑥 plus one, and this is equal to zero.
To solve this equation, we set each
factor equal to zero. This gives us two solutions: 𝑥
equals five and 𝑥 is equal to negative one. Whilst these are both valid roots
of the given quadratic equation, they are not both valid solutions to the
problem. The expression two 𝑥 represents
the length of the hypotenuse of the right triangle, and this must be strictly
positive. We can therefore disregard the
solution 𝑥 is equal to negative one, and the value of 𝑥 is five.
Whilst it is not required in this
question, we could substitute this value of 𝑥 back into the expressions for each of
the side lengths of the triangle. The right triangle has side lengths
six, eight, and 10. And since six squared plus eight
squared is equal to 10 squared, this is a right triangle.
Some equations we encounter may not
appear to be quadratic equations at first. But by making an appropriate
substitution, we see that they can be expressed as quadratic equations in another
variable. We will now consider one final
example in which we solve a quartic equation by recognizing that it can be expressed
as a quadratic equation.
Find the solution set for 𝑥 to the
fourth power minus 13𝑥 squared plus 36 is equal to zero.
The equation we’ve been asked to
solve is of degree four, and hence it is a quartic equation. However, by the laws of exponents
or indices, we know that 𝑥 to the fourth power is equal to 𝑥 squared squared. This means that we have a
quadratic-like equation. This can be rewritten as 𝑥 squared
squared minus 13 multiplied by 𝑥 squared plus 36 is equal to zero. We can then introduce a new
variable 𝑦 such that 𝑦 is equal to 𝑥 squared, and our equation becomes 𝑦 squared
minus 13𝑦 plus 36 equals zero.
We can now solve this equation for
𝑦 and then subsequently solve for 𝑥. The quadratic expression factors
into 𝑦 minus four multiplied by 𝑦 minus nine, as negative four and negative nine
have a sum of negative 13 and a product of 36. We therefore have two possible
solutions: 𝑦 minus four equals zero or 𝑦 minus nine equals zero. And hence 𝑦 is equal to four or 𝑦
is equal to nine.
Recalling our substitution and
solving for 𝑥, we have 𝑥 squared is equal to four and 𝑥 squared is equal to
nine. Both of these equations have two
real roots: 𝑥 is equal to positive or negative root four and 𝑥 is equal to
positive and negative root nine. Since the square root of four is
two and the square root of nine is three, we have 𝑥 is equal to positive or
negative two and positive or negative three. The solution set for 𝑥 to the
fourth power minus 13𝑥 squared plus 36 equals zero has four values: two, negative
two, three, and negative three.
We will now summarize the key
points from this video. A quadratic equation can be
expressed in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and
𝑐 are constants and 𝑎 is nonzero. To solve a quadratic equation of
this type, we follow three steps. We begin by expressing the equation
in factored form. Then, we set each factor equal to
zero and finally solve the resulting linear equations.
We saw in this video that some
nonquadratic equations can be solved by first factoring by a highest common factor
or by making an appropriate substitution to a quadratic-like equation. We also saw that worded and
practical problems can be solved by first forming a quadratic equation and then
solving by factoring. In this type of question, we will
sometimes need to disregard one or more solutions.
