### Video Transcript

The sum of two numbers is 11 and
the sum of their squares is 65. What are the numbers?

Now, itβs important to notice that
we shouldnβt answer this question using trial and error. We need to use a formal
approach. Weβre going to use some algebra to
answer the problem. So weβll let the two numbers be
represented by the letters π₯ and π¦. Weβre now going to express the
information in the question as equations involving π₯ and π¦. Firstly, the sum of the two numbers
is 11. So this gives us the equation π₯
plus π¦ is equal to 11.

Secondly, weβre told that the sum
of their squares is 65. So this gives us the equation π₯
squared plus π¦ squared equals 65. We now have a linear-quadratic
system of equations. The first equation is linear, and
the second is quadratic. Weβre going to use the method of
substitution to solve these two equations simultaneously. We begin by rearranging the linear
equation to give one variable in terms of the other. And in this problem, it doesnβt
make any difference which variable we choose because the problem is equally
complicated or equally simple in both variables. Iβve chosen to rewrite equation one
as π¦ equals 11 minus π₯.

We now take this expression for π¦
in terms of π₯ and substitute it into the second equation, thatβs into the quadratic
equation. Giving π₯ squared plus 11 minus π₯
all squared is equal to 65. And now, we have a quadratic
equation in one variable π₯ only. We can distribute the parentheses
carefully, remembering that 11 minus π₯ all squared means 11 minus π₯ multiplied by
11 minus π₯, which simplifies to 121 minus 22π₯ plus π₯ squared. We can group the like terms on the
left-hand side β thatβs π₯ squared plus π₯ squared β giving two π₯ squared. And at the same time, subtract 65
from each side to give the quadratic equation two π₯ squared minus 22π₯ plus 56 is
equal to zero.

Now, we notice at this point that
each of the coefficients in our equation are even numbers. So we can simplify by dividing the
entire equation by two. Doing so gives the simplified
quadratic equation π₯ squared minus 11π₯ plus 28 is equal to zero. Now, we want to solve this equation
for π₯. So we first look to see whether
this equation can be factored. As the coefficient of π₯ squared in
our equation is one, the first term in each set of parentheses will simply be π₯
because π₯ multiplied by π₯ gives π₯ squared. And to complete the parentheses,
weβre then looking for two numbers whose sum is the coefficient of π₯ β thatβs
negative 11 β and whose product is the constant term β thatβs 28.

By considering the factors of 28,
we see that if we choose the two numbers negative seven and negative four, then
their product is indeed 28 because a negative multiplied by a negative gives a
positive. And their sum is indeed negative
11. So these are the two numbers weβre
looking for to complete our parentheses. We have our quadratic in its
factored form then. π₯ minus seven multiplied by π₯
minus four is equal to zero.

To solve, we take each factor in
turn, set it equal to zero, and solve the resulting equation. We have π₯ minus seven equals zero,
which can be solved by adding seven to each side to give π₯ equals seven, and then
π₯ minus four equals zero, which we solve by adding four to each side to give π₯
equals four. We found then the solution for
π₯. There are two possible values. π₯ is either equal to four or π₯ is
equal to seven.

We now need to find the
corresponding π¦-values, which we can do by substituting each π₯-value in turn into
the linear equation which, remember, that was π¦ is equal to 11 minus π₯. When π₯ is equal to seven, we find
that π¦ is equal to 11 minus seven which is four. And when π₯ is equal to four, we
find that π¦ is equal to 11 minus four, which is seven. So we have exactly the same values
for π¦, as we did for π₯.

The reason for this is because at
the beginning of the problem, we just let the two numbers be represented by π₯ and
π¦. We didnβt specify whether π₯ or π¦
was the larger number. So we found the same solution
twice. The two numbers are seven and
four. And either can be π₯ and then the
other will be π¦. We can of course check our answer,
firstly, by confirming that the sum of our two numbers is 11, which of course it is,
and, secondly, by confirming that the sum of their squares β seven squared plus four
squared, which is 49 plus 16 β is indeed equal to 65.

So by first formulating the
information in the question as a linear-quadratic system of equations and then
solving this pair of simultaneous equations using the substitution method, we found
that the two numbers weβre looking for are seven and four.