### Video Transcript

In this lesson, weβll learn how to
recognize the relationship between the coefficients of a quadratic equation and its
roots. Remember, a quadratic function is
of the form π of π₯ equals ππ₯ squared plus ππ₯ plus π, where π, π, and π are
real numbers and π is not equal to zero.

Letβs consider a quadratic
equation. Now, this is generally of the form
ππ₯ squared plus ππ₯ plus π equals zero. Letβs look at the equation π₯
squared minus seven π₯ plus 10 equals zero. We can solve this equation, in
other words, find its roots, by factoring the left-hand side. And since the coefficient of π₯
squared is equal to one, this is fairly straightforward. We look for two numbers whose
product is 10 and whose sum is negative seven.

Well, those two numbers are
negative two and negative five. So the left-hand side can be
factored to π₯ minus two times π₯ minus five. Then, since the product of the two
expressions in parentheses is zero, we can say that either one of those expressions
must itself be equal to zero. In other words, π₯ minus two is
equal to zero or π₯ minus five is equal to zero. Solving each equation for π₯ gives
us two solutions to our equation: π₯ equals two and π₯ equals five.

But letβs think about the
factorization process in reverse. We notice that the roots of the
negative of the two numbers that we chose had a product of 10 and a sum of negative
seven. So if we define the roots to be π₯
sub one and π₯ sub two, we can say that negative π₯ sub one times negative π₯ sub
two is equal to 10 and negative π₯ sub one plus negative π₯ sub two is equal to
negative seven. In other words, weβre beginning
with a presumption that the coefficients of the quadratic equation actually contain
information about its roots. And we can verify that this is true
in our example, since our roots are π₯ sub one equals two and π₯ sub two equals
five. Negative two times negative five is
10, and negative two plus negative five is negative seven.

But is this true for any quadratic
equation? What is the relationship between
the coefficients of any quadratic equation and its roots? Letβs start with a simple quadratic
equation, a more general form of the one we just saw. π₯ squared plus ππ₯ plus π equals
zero. Say that this factors into some
expression π₯ minus π₯ sub one times π₯ minus π₯ sub two. Then, the two roots of our
quadratic equation are π₯ sub one and π₯ sub two. But say we were going to distribute
the parentheses. We would get π₯ squared minus π₯
sub two π₯ minus π₯ sub one π₯ plus π₯ sub one π₯ sub two equals zero. And we could factor the middle two
terms to get negative π₯ times π₯ sub one plus π₯ sub two.

Now, letβs compare the coefficients
of this equation with that in our original equation. We can say that π is equal to π₯
sub one π₯ sub two and π is equal to the negative sum of π₯ sub one and π₯ sub
two. In other words, quadratic equations
of the form weβve given with roots π₯ sub one and π₯ sub two must satisfy π₯ sub one
plus π₯ sub two equals negative π and π₯ sub one times π₯ sub two equals π.

Letβs generalize this further,
thinking about quadratic equations where their leading coefficient is not equal to
one. We can manipulate these to make the
coefficient of π₯ squared one. So we get π₯ squared plus π over
ππ₯ plus π over π equals zero. Weβve just divided through by
π. We can now use our previous results
to link the coefficients and roots of any quadratic equation. For a quadratic equation ππ₯
squared plus ππ₯ plus π equals zero whose roots are π₯ sub one and π₯ sub two,
they must satisfy the two equations π₯ sub one plus π₯ sub two equals negative π
over π and π₯ sub one times π₯ sub two equals π over π. And of course for simple quadratic
equations where the coefficient of π₯ squared is one, this can be simplified as
shown.

Whatβs really useful is these
formula hold for all quadratic equations, even when the roots are complex or are
repeated. And the same formulae can be
recovered using the quadratic formula. In our first example, weβll
demonstrate how this theorem can help us to find the sum of the roots without
actually solving an equation.

Without solving the equation
negative three π₯ squared minus 16π₯ plus 63 equals zero, find the sum of its
roots.

Remember, if weβre given a
quadratic equation of the form ππ₯ squared plus ππ₯ plus π equals zero with roots
π₯ sub one and π₯ sub two, these must satisfy the criteria that their sum is equal
to negative π over π and their product is equal to π over π. And so in order to be able to find
the sum of the roots of our quadratic equation, we need to identify the values of π
and π.

Well, if we look carefully, we can
see that the coefficient of π₯ squared gives us our value of π. Itβs negative three. The coefficient of π₯ gives us our
value of π. Thatβs negative 16. And π is the constant in our
equation. Itβs 63. Weβre looking to find the sum of
the roots of this equation. So if we define them, as in our
general formula, as π₯ sub one and π₯ sub two, we know that itβs equal to negative
π divided by π. In this case, thatβs negative
negative 16 divided by negative three. A negative divided by a negative is
a positive. So we get 16 over three, and then
we make that negative. And so π₯ sub one plus π₯ sub two
is negative 16 over three. And we can now see that, without
solving the quadratic equation, we found the sum of its roots. Itβs negative 16 over three.

Now, in fact, weβve demonstrated
how to use an equation to find information about its roots. But we can reverse this
process. In other words, if weβre given the
roots of an equation, we can use the relationship between the coefficient and their
roots to find the original equation itself. In this case, itβs generally
sensible to start with the most simple form where the coefficient of π₯ squared is
equal to one, so the equation π₯ squared plus ππ₯ plus π equals zero. Then, if either of the final
coefficients π and π is a fraction, we can simply multiply both sides of the
equation by a common denominator to further simplify it. Letβs demonstrate this in our next
example.

Given that negative one and
negative six are the solutions of the equation π₯ squared plus ππ₯ plus π equals
zero, find the values of π and π.

We might recall that, for a
quadratic equation of the form π₯ squared plus ππ₯ plus π equals zero, which has
roots π₯ sub one and π₯ sub two, the sum of these roots is equal to negative π and
their product is equal to π. Now, weβre told that the roots of
the equation are negative one and negative six. So letβs define π₯ sub one to be
equal to negative one and π₯ sub two to be equal to negative six. We know the sum of these two values
gives us negative π. So we have negative one plus
negative six equals negative π. Therefore, negative seven is equal
to negative π, which means π is equal to seven.

We can perform a similar process to
find the value of π. This time, itβs the product of the
roots. Itβs negative one times negative
six. Well, negative one times negative
six is simply six. So π must be equal to six. And whilst not required of us, we
could use this information to construct the original quadratic equation. Itβs π₯ squared plus seven π₯ plus
six equals zero.

Now, in fact, we can even check our
results by solving this equation. If we factor the left-hand side, we
get π₯ plus one times π₯ plus six, which does give us the two solutions and the two
roots negative one and negative six, as required. So π is equal to seven and π is
equal to six.

Now, the beauty of what weβve
demonstrated is that it holds for all types of roots. Specifically, it even holds for
complex roots. So letβs see what that might look
like.

Which quadratic equation has roots
π₯ is equal to two plus or minus π? Is it (A) π₯ squared minus four π₯
plus five equals zero? (B) π₯ squared plus four π₯ plus
five equals zero. Is it (C) π₯ squared minus four π₯
plus three equals zero? (D) π₯ squared plus four π₯ plus
three equals zero. Or (E) π₯ squared minus five π₯
plus four equals zero.

Now, we could use some methods such
as completing the square or the quadratic formula to solve each of these individual
equations. But remember, we have a way to link
the coefficients of a quadratic equation with its roots. Specifically, letβs think about a
quadratic equation with a coefficient of π₯ squared is equal to one. Letβs say that equation has roots
π₯ sub one and π₯ sub two and the coefficient of π₯ is π and the constant term is
π. The sum of the roots is equal to
negative π, and the product of the roots is equal to π.

So, given information about our two
roots, we can find the value of negative π by adding them and we can find the value
of π by multiplying them. Specifically, negative π is two
plus π plus two minus π. And of course we can add pairs of
complex numbers by just adding the real parts and then separately adding the
imaginary parts. So thatβs two plus two plus π
minus π. But π minus π is simply zero, so
that gives us a real value of four. And since negative π is equal to
four, we can say π, which is going to be the coefficient of π₯ in our equation,
must be equal to negative four. And this is great, in fact, because
we can instantly disregard option (B) and (D) from our solutions. They have a coefficient of π₯ of
positive four.

Next, weβll find the product of the
roots to find the value of π. Thatβs two plus π times two minus
π. Distributing the parentheses, and
we get four minus two π plus two π minus π squared. Negative two π plus two π is
zero. And since π squared is negative
one, this simplifies further to four minus negative one, which is simply five. So we have a quadratic equation
where the coefficient of π₯ squared, we said, is one; the coefficient of π₯, π, is
negative four; and the constant term π is five. So our equation is π₯ squared minus
four π₯ plus five equals zero. And that, we see, corresponds to
answer (A).

So now that weβve looked at several
examples that demonstrate the link between the coefficients and roots of a quadratic
equation, letβs look at how we can solve more complicated problems involving
unknowns.

If πΏ and π are the roots of the
equation π₯ squared plus 10π₯ plus nine equals zero, what is the value of πΏ squared
plus π squared?

Now, one way we can answer this
problem would be to solve the original equation. But we are in fact given
information about the roots, and weβre given the equation itself. And so we recall that, given a
quadratic equation with a coefficient of π₯ squared is one, and if it has roots π₯
sub one and π₯ sub two, the sum of these roots is negative π and the product of
these roots is π. So our roots are πΏ plus π. So whatβs their sum?

Well, the coefficient of π₯ here,
the value of π, is 10. So πΏ plus π must be equal to
negative 10. Similarly, the product of these
roots must be equal to the constant term, which is nine. So we have two equations: πΏ plus
π equals negative 10 and πΏπ equals nine.

And we might not know where to go
next, but we are trying to find the value of πΏ squared plus π squared. So letβs imagine what might happen
if we square both sides of this first equation. When we do, we get πΏ plus π all
squared equals negative 10 squared. But then if we distribute the
left-hand side, we get the expression πΏ squared plus two πΏπ plus π squared. And we can see that that
corresponds in some respects to the value that weβre looking for. And in fact negative 10 squared is
100. So this is equal to 100.

Since we have two πΏπ and we know
πΏπ is equal to nine, we can rewrite two πΏπ as two times nine, which is 18. And so we have an equation πΏ
squared plus 18 plus π squared equals 100. Then, we solve for πΏ squared plus
π squared by subtracting 18 from both sides. 100 minus 18 of course is 82. And so the value of πΏ squared plus
π squared is 82.

Weβve now demonstrated in some
detail how coefficients of a quadratic equation contain information about the sum
and product of its roots. Letβs recap the key concepts.

In this lesson, we learned that,
for a quadratic equation ππ₯ squared plus ππ₯ plus π equals zero, if it has roots
π₯ sub one and π₯ sub two, then the sum of these values are negative π over π. And similarly, the product of the
roots is π over π. Of course, if π is equal to one,
if the coefficient of π₯ squared equals one, then this simplifies to π₯ sub one plus
π₯ sub two equals negative π and π₯ sub one times π₯ sub two equals π. And of course we can reverse this
idea. We can use information about the
roots to find coefficients of a quadratic equation and, hence, find the equation
itself.