### Video Transcript

In this lesson, weβll learn how to
write a quadratic equation given the roots of another quadratic equation. Letβs begin by recalling the
relationship between the roots of a quadratic and the coefficients of its terms.

For a quadratic equation of the
form ππ₯ squared plus ππ₯ plus π equals zero, by the quadratic formula, its
solutions, and as its roots are π one and π two, are π one equals negative π
plus the square root of π squared minus four ππ over two π. And π two is equal to negative π
minus the square root of π squared minus four ππ over two π. Weβre now going to find the sum and
product of these general roots.

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

Weβre now going to repeat this
process to find the product. Product means multiply, so weβre
doing π one times π two. Once again, we split the first
fraction up, and then we split the second fraction up. And weβre 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 π squared over four π squared. Then, multiplying the outer and
inner terms gives us the positive and negative versions of π times the square root
of π squared minus four ππ all over four π 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 π squared minus four ππ all over four π squared. Weβre now going to split this
second fraction up, remembering, of course, that since weβre going to be subtracting
negative four ππ over four π squared, thatβs the same as adding it. And so we should see now that π
squared over four π squared minus π squared over four π squared is zero. Similarly, we can divide through by
an π and by the four in this final fraction, and that leaves us simply with π over
π. 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 π₯ squared.

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

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

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

Letβs 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 π₯ squared minus that sum of the roots times π₯ plus the product of the roots
equals zero. And so, if we consider our equation
π₯ squared plus eight π₯ plus 12 equals zero, the sum of the roots must be negative
eight. Now, the reason itβs negative eight
and not positive eight is because in the general form, the coefficient of π₯ 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 weβll look to replace the
roots with the roots given in the question. So here the roots are πΏ plus three
and π plus three. But in fact, weβre 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 π
one and π two, we see that theyβre equal to negative six and negative two.

But of course, we were told that
the roots of this equation are πΏ plus three and π plus three. So we can form two equations, one
in πΏ and one in π. The first equation is πΏ plus three
equals negative six, and the second is π plus three equals negative two. Weβll solve both of these equations
for their respective variable by subtracting three from both sides. Negative six minus three is
negative nine. So πΏ must be equal to negative
nine. Similarly, π 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 π₯ squared minus
negative 14 times π₯ plus 45 equals zero, and this simplifies to π₯ squared plus
14π₯ plus 45 equals zero.

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

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

Letβs begin by recalling the
relationship between a quadratic equation whose leading coefficient is one and its
roots. We can represent it as π₯ squared
minus the sum of the roots times π₯ 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 π₯ squared is one, the negative coefficient
of π₯ will tell us the sum of the roots and the constant term will tell us its
product.

So we take the equation π₯ squared
minus two π₯ plus five equals zero. The coefficient of π₯ 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. Weβre not going to get nice integer
solutions. And so instead, weβre going to form
equations using πΏ and π. Since the sum of our roots is two
and πΏ and π are the roots, we can say πΏ plus π must be equal to two. And then we can say that πΏ times
π is equal to five.

The roots of our new equation are
πΏ squared and π squared. And so, since their sum will be πΏ
squared plus π squared, we need to manipulate our equations to find an expression
for πΏ squared plus π squared and another expression for their product, πΏ squared
π squared. Letβs label our equations as one
and two. Weβre going to take the entirety of
equation one and, weβre 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
πΏ plus π squared, which we can consider to be πΏ plus π times πΏ plus π.

And if we distribute these
parentheses, we get πΏ squared plus two πΏπ plus π squared equals four. And then, if we subtract two πΏπ
from both sides, we get the expression for the sum of the roots πΏ squared and π
squared. Itβs four minus two πΏπ. But of course, we have an
expression for πΏπ; equation two tells us that πΏπ is equal to five. And so, πΏ squared plus π 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 π₯.

Weβre now going to repeat this
process for equation two; weβre going to square both sides. That is, πΏπ 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 πΏ squared π squared
equals 25. We, therefore, find that the sum of
our new roots is negative six and the product is 25. Letβs substitute these back into
the general form. When we do, we get π₯ squared minus
negative six π₯ plus 25 equals zero. And so, the quadratic equation
whose roots are πΏ squared and π squared is π₯ squared plus six π₯ plus 25 equals
zero.

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

If πΏ and π are the roots of the
equation π₯ squared plus 20π₯ plus 15 equals zero, what is the value of one over π
plus one over πΏ?

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 π₯ squared is one,
the negative coefficient of π₯ tells us the sum of the roots and the constant term
tells us the product. And this is really useful because
the coefficient of π₯ here is 20 and the constant is 15. And so, the sum of our roots must
be negative 20. Remember, we said that itβs the
negative coefficient of π₯. Then, the product, which is the
constant term, must be 15.

But of course, we were told that πΏ
and π are the roots of our equation. So we can, in fact, say that πΏ
plus π must be negative 20 and πΏ times π must be 15. So, how does this help? Weβre looking to find the value of
one over π plus one over πΏ, and we canβt easily find two numbers that have a sum
of negative 20 and a product of 15. So weβre going to need to
manipulate our expressions. Letβs think of one over π plus one
over πΏ.

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, weβre going to multiply the
numerator and denominator of our first fraction by πΏ and of our second fraction by
π. That gives us πΏ over πΏπ plus π
over πΏπ. And now since the denominators are
the same, we simply add the numerators, and that gives us πΏ plus π over πΏπ.

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

Weβre now going to extend this idea
into forming quadratic equations.

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

Letβs 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 π₯ 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 π₯ squared, is three. And so, weβre going to divide every
single term by three. Three π₯ squared divided by three
is π₯ squared. 16π₯ divided by three is 16π₯ over
three or 16 over three π₯. 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 π₯. So, itβs negative 16 over
three. Then, the product is negative
one-third. But weβre also told that πΏ and π
are the roots of our equation, so we can replace these with πΏ plus π as the sum
and πΏ times π as the product. Weβre looking to find the quadratic
equation whose roots are πΏ over two and π over two. And so, what weβre going to do is
find an expression for the sum of these roots; thatβs πΏ over two plus π over
two. Similarly, weβre going to find the
product of these roots; thatβs πΏ over two times π over two which is πΏπ over
four.

So, how can we link the equations
we have? Well, letβs call this first
equation one. We have πΏ plus π as being equal
to negative 16 over three. If we divide the entire expression
by two, that is πΏ plus π over two, we actually know thatβs equal to πΏ over two
plus π over two. And so, this means we can find the
value of πΏ over two plus π over two by dividing the value for πΏ plus π by
two. Thatβs 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 πΏπ
divided by four. So thatβs 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 πΏ over two and π over two is π₯ squared minus
negative eight over three π₯ plus negative one twelfth equals zero.

Letβs 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, weβre going to
multiply through by 12. π₯ squared times 12 is 12π₯
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
12π₯ squared plus 32π₯ minus one equals zero.

Letβs now recap the key points from
this lesson. In this video, we learned that for
an equation of the form ππ₯ squared plus ππ₯ plus π equals zero, the sum of the
roots is negative π over π. Itβs the negative coefficient of π₯
divided by the coefficient of π₯ squared. And the product is π over π. Itβs the constant term divided by
the coefficient of π₯ 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 π₯ tells us the sum of the roots, whereas the
constant term tells us the product.