Quadratic Equations (Level 3) In the previous video we went over how to
solve quadratic monomials of the form ax squared equals 0 and quadratic binomials of the form
ax squared plus c equals 0, now we will learn how to solve a slightly more complicated version
of these quadratic equations specifically quadratic equations that contain the sum of
a variable and a constant squared in other words quadratic equations that contain the
square of a binomial of the form a times the quantity px plus q, squared equals 0 and a
times the quantity px plus q, squared plus c equals 0. These types of quadratic equations are
solved similarly like the equations that we solved from the previous video we can solve
them by taking the square root of both sides of the equation. Alright let’s start with
the first example. Solve the equation the quantity 3x plus 1, squared equals
0. Notice that this quadratic equation contains
a binomial squared so we can easily solve this equation by taking the square root of
both sides so we do just that and obtain the following, so here remember that the square
root of 3x + 1 squared reduces to the absolute value of 3x + 1 because this is the definition
of the absolute value, remember the square root of a variable squared is equal to its
absolute value. Next we take the square root of 0 which reduces to zero, and now we need
to the solve the following absolute value equation, this equation is equivalent to the
linear equation 3x plus 1 equals zero so we subtract 1 to both sides and obtain the following
3x equals negative 1 and finally we divide both sides of the equation by 3 and obtain
the final answer x = negative 1 third. Notice that we obtained one answer, recall that we
always need to obtain two answers when solving quadratic equations, if we graph the left
hand side of the equation we would obtain the following graph, notice that the graph
touches the x-axis at negative (-1/3, 0), only once, this is an example of a repeated or
multiple root of a quadratic equation, more about this in a later video, let’s move
along and try the next example.
Solve the equation the quantity x minus 1 squared minus 36 equals 0.
First step in solving any polynomial is to write it in standard form, in this case the
equation is already in standard form so we start by moving the constant to the other
side by adding 36 to both sides of the equation as follows, doing that we obtain the following,
next we take the square root of both sides of the equation, the left hand side simplifies
to the absolute value of x - 1 and the right side simplifies to 6, next we solve the absolute
value equation, recall that there are two answers and as a result we need to set up two
distinct equations. One of them is x -1 equals 6 and the other is x - 1 = -6, so we solve
each equation individually and obtain x = 7 as one solution and x = -5 as the other solution.
Once again notice that we ended up with two answers as expected from quadratic equations.
Recall that in the previous video we skipped the absolute value steps and went straight
to the answer by adding the positive negative symbol immediately after taking the square
root of a number, from this point on I will do just that and write the answer using positive
negative notation as follows, x = 1 positive negative 6, note that whenever possible simplify
the expression so that you explicitly show both answers for example this expression can
be simplified as x = 1 + 6 and x = 1 - 6 which simplifies to 7 and negative 5 respectively.
It’s also a good idea to check your answers especially during quizzes and tests.
Let’s go ahead and try the next equation. Solve the equation 5 times the quantity z plus 2 squared
minus 125 equals 0. First step is to write it in standard form luckily
it’s all ready written for us. Next we need to get the constant to the other side of
the equation, so we add 125 to both sides. Next we need to get rid of 5, in this case
notice that 5 is being multiplied by the binomial squared so we first need to take care of the
5 before we take the square root of both sides, so dividing both sides of the equation by
5 we obtain the quantity z + 2 squared = 25, the next step is to take the square root of
both sides doing that we obtain the following z plus 2 = positive negative 5, then we solve
for z by subtracting 2 from both sides as follows, then it’s just a matter of finding
both solutions so we need to break apart this expression one expression would be obtained by using the positive
operator and the other by using the negative operator then we simplify to obtain 3 and
negative 7 these numbers represent the zeros or roots of the quadratic equation.
Let’s try the next one. Solve the equation the quantity x minus 2/5 squared
equals 9 over 25. Notice that we have fractions in this problem
so don’t freak out for the most part many students don’t like fractions and tend to
run away from problems that contain them. They are really not that bad, it’s just
a matter of remembering how to manipulate them. Notice that this problem is
set up ready for us to take the square root of both sides so we do just that. The left
hand side simplifies to x minus two fifths, and the right hand side can be simplified
by taking the square root of both the numerator and denominator as follows and this expression
simplifies to positive negative three fifths. Next we solve for x by adding two fifths to
both sides. Now we need to add and subtract fractions, remember the way you add or subtract
fraction is by first modifying the fractions so that they contain a common denominator
then it’s just a matter of adding or subtracting the numerator. In this case both fractions
have a common denominator so we just add the numerators to obtain 5 over 5 which simplifies
to 1 and for the second expression we have two fifths minus three fifths which simplifies to
negative one fifth.
Alright let’s try the next example. Solve the equation the quantity y minus 2 squared minus
24 equals 0. This problem is already in standard form so
we add 24 to both sides of the equation, next we take the square root of both sides as follows
the left hand side simplifies to y - 2 and the right hand side simplifies to positive
negative 2 root 6, remember the way you simplify radical expressions is to first break down
the number inside radicand to its prime factorization, then separate any perfect squares in this
case we have 2 squared which is equal to 4 which is a perfect square, separating it and taking
the square root we obtain positive negative 2 square root of 6. Finally we solve for y by
adding two to both sides of the equation and obtain the following expression. 2 positive
negative 2 square root of 6. Notice that we can’t simplify this expression like we did
in the previous examples; so we leave our answer in this exact form as oppose to a decimal
form. Notice that we ended up with a combination of rational and irrational numbers for our
answer, keep in mind that these types of answers can and will occur when dealing with quadratic
equations. Let’s go ahead and try the next example.
Solve the equation 3 times the quantity x minus 4 squared plus 12 equals 0.
Alright this equation is currently in standard form, so we first subtract 12 from both sides
as follows, next we divide both sides of the equation by 3 and obtain the following expression,
x - 4 squared = -4, now we take the square root of both sides, doing that we obtain the
following, here the square root of negative 4 reduces to positive negative 2i, remember
the square root of a negative number is equal to i times the square root of that number.
The last step is to solve for x by adding 4 to both sides of the equation. So our final
answer is 4 plus 2i and 4 minus 2i, notice that we ended up with complex numbers as our
solutions, remember quadratic equations may have real, imaginary, or complex solutions.
Now let’s end this video with a final example. Solve the equation one half times the quantity 4x plus
3 squared plus 13 equals 3. This problem requires nearly all the techniques
that we have seen from the previous examples, so let’s start by solving for x by subtracting
13 from both sides of the equation, simplifying the expressions we obtain the following. Next
we multiply both sides of the equation by 2 to get rid of the fraction one half, doing
that we have the following, next we go ahead and take the square root of both sides as
follows here the square root of negative 20 simplifies to 2i root 5 or it can also written
as 2 times root 5 times i, then we simplify this expression as follows, next step is to subtract 3 from
both sides, and finally we solve for x by dividing both sides of the equation by
4, doing that we obtain the following the quantity negative 3 positive negative 2i times the square root
of 5 all over 4, we can also divide 4 to each term as follows, and simplify the second term
as follows, keep in mind that either form of the solution is acceptable. So it turns
out that the solution to this equation is equal to negative ¾ + i times the square root of 5 over 2,
and negative ¾ minus i times square root of 5 over 2, and these crazy looking expressions
are the solutions of this particular quadratic equation.
Alright in our next video we will learn how to solve specific quadratic equations by factoring.