### Video Transcript

In this lesson, we’ll learn how to
write a quadratic equation given the roots of another quadratic equation. Let’s begin by recalling the
relationship between the roots of a quadratic and the coefficients of its terms.

For a quadratic equation of the
form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, by the quadratic formula, its
solutions, and as its roots are 𝑟 one and 𝑟 two, are 𝑟 one equals negative 𝑏
plus the square root of 𝑏 squared minus four 𝑎𝑐 over two 𝑎. And 𝑟 two is equal to negative 𝑏
minus the square root of 𝑏 squared minus four 𝑎𝑐 over two 𝑎. We’re now going to find the sum and
product of these general roots.

The sum is 𝑟 one plus 𝑟 two. The expression for the sum is
shown. But of course, we can split the
first fraction up into negative 𝑏 over two 𝑎 plus the square root of 𝑏 squared
minus four 𝑎𝑐 over two 𝑎. And then, we can split the second
fraction into negative 𝑏 over two 𝑎 minus the square root of 𝑏 squared minus four
𝑎𝑐 over two 𝑎. And now, we see that the second
term and the fourth term sum to make zero; essentially, they cancel. And so, we have negative 𝑏 over
two 𝑎 plus negative 𝑏 over two 𝑎, which is negative two 𝑏 over two 𝑎. Finally, we can cancel by dividing
both the numerator and denominator of our fraction by two, leaving us with negative
𝑏 over 𝑎. So, the sum of our roots is
negative 𝑏 over 𝑎. In other words, it’s the negative
of the coefficient of 𝑥 over the coefficient of 𝑥 squared.

We’re now going to repeat this
process to find the product. Product means multiply, so we’re
doing 𝑟 one times 𝑟 two. Once again, we split the first
fraction up, and then we split the second fraction up. And we’re now going to distribute
by multiplying the first terms, the outer terms, the inner terms, and the last
terms. Multiplying the first terms gives
us 𝑏 squared over four 𝑎 squared. Then, multiplying the outer and
inner terms gives us the positive and negative versions of 𝑏 times the square root
of 𝑏 squared minus four 𝑎𝑐 all over four 𝑎 squared. Now, of course, the sum of these is
zero. So, these terms are actually going
to cancel.

Then, when we multiply the last
terms, we get negative 𝑏 squared minus four 𝑎𝑐 all over four 𝑎 squared. We’re now going to split this
second fraction up, remembering, of course, that since we’re going to be subtracting
negative four 𝑎𝑐 over four 𝑎 squared, that’s the same as adding it. And so we should see now that 𝑏
squared over four 𝑎 squared minus 𝑏 squared over four 𝑎 squared is zero. Similarly, we can divide through by
an 𝑎 and by the four in this final fraction, and that leaves us simply with 𝑐 over
𝑎. And we can therefore say that the
product of the two roots of a quadratic equation is the constant term divided by the
coefficient of 𝑥 squared.

Now whilst this might not seem like
a hugely useful result, if we go back to our original equation 𝑎𝑥 squared plus
𝑏𝑥 plus 𝑐 equals zero and divide through by 𝑎 — we’re allowed to do this, of
course, cause 𝑎 is not going to be equal to zero — we see that our quadratic
equation becomes 𝑥 squared plus 𝑏 over 𝑎𝑥 plus 𝑐 over 𝑎 equals zero. And now, we should see that these
link to our earlier results. Negative 𝑏 over 𝑎 is the same as
the negative coefficient of 𝑥, and 𝑐 over 𝑎 is the constant term. And so, we’re now able to say that
a quadratic equation whose leading coefficient is one can be written as 𝑥 squared
minus the sum of roots times 𝑥 plus the product of the roots equals zero.

So, let’s have a look at an example
of how we might apply these results.

Given that 𝐿 plus three and 𝑚
plus three are the roots of the equation 𝑥 squared plus eight 𝑥 plus 12 equals
zero, find, in its simplest form, the quadratic equation whose roots are 𝐿 and
𝑚.

Let’s begin by recalling the
relationship between a quadratic equation whose leading coefficient is one and its
roots. We can write the equation in the
form 𝑥 squared minus that sum of the roots times 𝑥 plus the product of the roots
equals zero. And so, if we consider our equation
𝑥 squared plus eight 𝑥 plus 12 equals zero, the sum of the roots must be negative
eight. Now, the reason it’s negative eight
and not positive eight is because in the general form, the coefficient of 𝑥 is the
negative sum of the roots, and in our example, the coefficient is positive. We can also say that the product,
which is the constant term, must be equal to 12.

And often we’ll look to replace the
roots with the roots given in the question. So here the roots are 𝐿 plus three
and 𝑚 plus three. But in fact, we’re able to
calculate two numbers which have a sum of negative eight and a product of 12. Two numbers which satisfy this
result are negative six and negative two, and this is because a negative times a
negative is a positive. So, negative six times negative two
is positive 12. But also, negative six plus
negative two is negative eight. And so, defining our roots to be 𝑟
one and 𝑟 two, we see that they’re equal to negative six and negative two.

But of course, we were told that
the roots of this equation are 𝐿 plus three and 𝑚 plus three. So we can form two equations, one
in 𝐿 and one in 𝑚. The first equation is 𝐿 plus three
equals negative six, and the second is 𝑚 plus three equals negative two. We’ll solve both of these equations
for their respective variable by subtracting three from both sides. Negative six minus three is
negative nine. So 𝐿 must be equal to negative
nine. Similarly, 𝑚 is equal to negative
five. And so this is really useful
because we now know the roots to our new equation. We can form this equation by
finding the sum of these roots and the product. The sum is negative nine plus
negative five, which is negative 14. And the product is negative nine
times negative five, which is 45.

And so now that we know that the
sum of our roots is negative 14 and the product is 45, we can substitute these back
into the general form. That gives us 𝑥 squared minus
negative 14 times 𝑥 plus 45 equals zero, and this simplifies to 𝑥 squared plus
14𝑥 plus 45 equals zero.

We’re now going to look at a
similar example. But this time, we’re not going to
be able to easily work out the roots of our original equation, and so we’re going to
use algebra to find the new equation.

Given that 𝐿 and 𝑚 are the roots
of the equation 𝑥 squared minus two 𝑥 plus five equals zero, find, in its simplest
form, the quadratic equation whose roots are 𝐿 squared and 𝑚 squared.

Let’s begin by recalling the
relationship between a quadratic equation whose leading coefficient is one and its
roots. We can represent it as 𝑥 squared
minus the sum of the roots times 𝑥 plus the product of the roots equals zero. And so, this is essentially saying
that if we have a quadratic equation equal to zero and the leading coefficient is
one, in other words, the coefficient of 𝑥 squared is one, the negative coefficient
of 𝑥 will tell us the sum of the roots and the constant term will tell us its
product.

So we take the equation 𝑥 squared
minus two 𝑥 plus five equals zero. The coefficient of 𝑥 is negative
two. And so the sum must be the negative
of negative two, so the sum of the roots must be positive two. Then the constant term is five. So the product of the roots must be
five. So can we find two numbers that
have a sum of two and a product of five? Well, no, not easily. We’re not going to get nice integer
solutions. And so instead, we’re going to form
equations using 𝐿 and 𝑚. Since the sum of our roots is two
and 𝐿 and 𝑚 are the roots, we can say 𝐿 plus 𝑚 must be equal to two. And then we can say that 𝐿 times
𝑚 is equal to five.

The roots of our new equation are
𝐿 squared and 𝑚 squared. And so, since their sum will be 𝐿
squared plus 𝑚 squared, we need to manipulate our equations to find an expression
for 𝐿 squared plus 𝑚 squared and another expression for their product, 𝐿 squared
𝑚 squared. Let’s label our equations as one
and two. We’re going to take the entirety of
equation one and, we’re going to square it. In other words, we square both
sides. So on the right-hand side, we get
two squared, which is, of course, equal to four. Then, on the left-hand side, we get
𝐿 plus 𝑚 squared, which we can consider to be 𝐿 plus 𝑚 times 𝐿 plus 𝑚.

And if we distribute these
parentheses, we get 𝐿 squared plus two 𝐿𝑚 plus 𝑚 squared equals four. And then, if we subtract two 𝐿𝑚
from both sides, we get the expression for the sum of the roots 𝐿 squared and 𝑚
squared. It’s four minus two 𝐿𝑚. But of course, we have an
expression for 𝐿𝑚; equation two tells us that 𝐿𝑚 is equal to five. And so, 𝐿 squared plus 𝑚 squared
becomes four minus two times five, which is four minus 10 or simply negative
six. So, we found the sum of the roots
of our new equation and, therefore, the negative coefficient of 𝑥.

We’re now going to repeat this
process for equation two; we’re going to square both sides. That is, 𝐿𝑚 squared equals five
squared. But, of course, five squared is
25. And we can distribute to the power
of two across both terms. And we get 𝐿 squared 𝑚 squared
equals 25. We, therefore, find that the sum of
our new roots is negative six and the product is 25. Let’s substitute these back into
the general form. When we do, we get 𝑥 squared minus
negative six 𝑥 plus 25 equals zero. And so, the quadratic equation
whose roots are 𝐿 squared and 𝑚 squared is 𝑥 squared plus six 𝑥 plus 25 equals
zero.

In our next example, we’ll look at
using the relationship between the coefficients of a quadratic and its roots to help
us find the value of an expression.

If 𝐿 and 𝑚 are the roots of the
equation 𝑥 squared plus 20𝑥 plus 15 equals zero, what is the value of one over 𝑚
plus one over 𝐿?

We begin by reminding ourselves
what the relationship between the coefficient of a quadratic equation is and its
roots. For a quadratic equation whose
leading coefficient is one, in other words, the coefficient of 𝑥 squared is one,
the negative coefficient of 𝑥 tells us the sum of the roots and the constant term
tells us the product. And this is really useful because
the coefficient of 𝑥 here is 20 and the constant is 15. And so, the sum of our roots must
be negative 20. Remember, we said that it’s the
negative coefficient of 𝑥. Then, the product, which is the
constant term, must be 15.

But of course, we were told that 𝐿
and 𝑚 are the roots of our equation. So we can, in fact, say that 𝐿
plus 𝑚 must be negative 20 and 𝐿 times 𝑚 must be 15. So, how does this help? We’re looking to find the value of
one over 𝑚 plus one over 𝐿, and we can’t easily find two numbers that have a sum
of negative 20 and a product of 15. So we’re going to need to
manipulate our expressions. Let’s think of one over 𝑚 plus one
over 𝐿.

We know that to add two fractions,
we need to create a common denominator. Now, the easiest way to do this
when working with algebraic fractions is to multiply both parts of each fraction by
the denominator of the other. So, we’re going to multiply the
numerator and denominator of our first fraction by 𝐿 and of our second fraction by
𝑚. That gives us 𝐿 over 𝐿𝑚 plus 𝑚
over 𝐿𝑚. And now since the denominators are
the same, we simply add the numerators, and that gives us 𝐿 plus 𝑚 over 𝐿𝑚.

And this is really useful because
we know that the numerator 𝐿 plus 𝑚 is equal to negative 20, and then the
denominator 𝐿𝑚 is 15. And this means, in turn, that one
over 𝑚 plus one over 𝐿 is equal to negative 20 over 15, which simplifies to
negative four-thirds. And so, if 𝐿 and 𝑚 are the roots
of our equation, then one over 𝑚 plus one over 𝐿 must be equal to negative
four-thirds.

We’re now going to extend this idea
into forming quadratic equations.

Given that 𝐿 and 𝑚 are the roots
of the equation three 𝑥 squared plus 16𝑥 minus one equals zero, find, in its
simplest form, the quadratic equation whose roots are 𝐿 over two and 𝑚 over
two.

Let’s begin by recalling the
relationship between the quadratic equation and its roots. For a quadratic equation whose
leading coefficient is one, we can say that the coefficient of 𝑥 is the negative
sum of the roots of the equation, whereas the constant is the product of its
roots. Now, comparing this to our
equation, we see we do have a bit of a problem. The leading coefficient, the
coefficient of 𝑥 squared, is three. And so, we’re going to divide every
single term by three. Three 𝑥 squared divided by three
is 𝑥 squared. 16𝑥 divided by three is 16𝑥 over
three or 16 over three 𝑥. And then, our constant term becomes
negative one-third.

And so, comparing this to the
general form, we know that the sum of the roots of our equation is the negative
coefficient of 𝑥. So, it’s negative 16 over
three. Then, the product is negative
one-third. But we’re also told that 𝐿 and 𝑚
are the roots of our equation, so we can replace these with 𝐿 plus 𝑚 as the sum
and 𝐿 times 𝑚 as the product. We’re looking to find the quadratic
equation whose roots are 𝐿 over two and 𝑚 over two. And so, what we’re going to do is
find an expression for the sum of these roots; that’s 𝐿 over two plus 𝑚 over
two. Similarly, we’re going to find the
product of these roots; that’s 𝐿 over two times 𝑚 over two which is 𝐿𝑚 over
four.

So, how can we link the equations
we have? Well, let’s call this first
equation one. We have 𝐿 plus 𝑚 as being equal
to negative 16 over three. If we divide the entire expression
by two, that is 𝐿 plus 𝑚 over two, we actually know that’s equal to 𝐿 over two
plus 𝑚 over two. And so, this means we can find the
value of 𝐿 over two plus 𝑚 over two by dividing the value for 𝐿 plus 𝑚 by
two. That’s negative 16 over three
divided by two, which is negative eight over three.

And we can repeat this process with
our second equation. This time, of course, we want 𝐿𝑚
divided by four. So that’s going to be negative
one-third divided by four, which is negative one twelfth. And now that we have the sum of our
roots and the product, we can substitute these back into our earlier equation. When we do, we find that the
quadratic equation whose roots are 𝐿 over two and 𝑚 over two is 𝑥 squared minus
negative eight over three 𝑥 plus negative one twelfth equals zero.

Let’s simplify a little by dealing
with our signs. In other words, negative negative
eight-thirds is just eight-thirds. And adding negative one twelfth is
the same as subtracting one twelfth. Our final step is to create integer
coefficients. And to do so, we’re going to
multiply through by 12. 𝑥 squared times 12 is 12𝑥
squared. Then, if we multiply eight-thirds
by 12, we cancel a three. And we end up working out eight
times four, which is 32. Negative one twelfth times 12 is
negative one. And, of course, zero times 12 is
zero. And so, the quadratic equation is
12𝑥 squared plus 32𝑥 minus one equals zero.

Let’s now recap the key points from
this lesson. In this video, we learned that for
an equation of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, the sum of the
roots is negative 𝑏 over 𝑎. It’s the negative coefficient of 𝑥
divided by the coefficient of 𝑥 squared. And the product is 𝑐 over 𝑎. It’s the constant term divided by
the coefficient of 𝑥 squared. We can use this information to
express the relationship between the roots of a quadratic whose leading coefficient
is one and the coefficient of its terms. When the equation is written in
this form, the negative coefficient of 𝑥 tells us the sum of the roots, whereas the
constant term tells us the product.