Video Transcript
The sum of two numbers is 11
and the sum of their squares is 65. What are the numbers?
Now, it’s important to notice
that we shouldn’t answer this question using trial and error. We need to use a formal
approach. We’re going to use some algebra
to answer the problem. So we’ll let the two numbers be
represented by the letters 𝑥 and 𝑦. We’re now going to express the
information in the question as equations involving 𝑥 and 𝑦. Firstly, the sum of the two
numbers is 11. So this gives us the equation
𝑥 plus 𝑦 is equal to 11.
Secondly, we’re told that the
sum of their squares is 65. So this gives us the equation
𝑥 squared plus 𝑦 squared equals 65. We now have a linear-quadratic
system of equations. The first equation is linear,
and the second is quadratic. We’re 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 doesn’t
make any difference which variable we choose because the problem is equally
complicated or equally simple in both variables. I’ve chosen to rewrite equation
one as 𝑦 equals 11 minus 𝑥.
We now take this expression for
𝑦 in terms of 𝑥 and substitute it into the second equation, that’s into the
quadratic equation. Giving 𝑥 squared plus 11 minus
𝑥 all squared is equal to 65. And now, we have a quadratic
equation in one variable 𝑥 only. We can distribute the
parentheses carefully, remembering that 11 minus 𝑥 all squared means 11 minus
𝑥 multiplied by 11 minus 𝑥, which simplifies to 121 minus 22𝑥 plus 𝑥
squared. We can group the like terms on
the left-hand side — that’s 𝑥 squared plus 𝑥 squared — giving two 𝑥
squared. And at the same time, subtract
65 from each side to give the quadratic equation two 𝑥 squared minus 22𝑥 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 𝑥 squared minus 11𝑥 plus 28 is equal to zero. Now, we want to solve this
equation for 𝑥. So we first look to see whether
this equation can be factored. As the coefficient of 𝑥
squared in our equation is one, the first term in each set of parentheses will
simply be 𝑥 because 𝑥 multiplied by 𝑥 gives 𝑥 squared. And to complete the
parentheses, we’re then looking for two numbers whose sum is the coefficient of
𝑥 — that’s negative 11 — and whose product is the constant term — that’s
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
we’re looking for to complete our parentheses. We have our quadratic in its
factored form then. 𝑥 minus seven multiplied by 𝑥
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 𝑥 minus seven equals
zero, which can be solved by adding seven to each side to give 𝑥 equals seven,
and then 𝑥 minus four equals zero, which we solve by adding four to each side
to give 𝑥 equals four. We found then the solution for
𝑥. There are two possible
values. 𝑥 is either equal to four or
𝑥 is equal to seven.
We now need to find the
corresponding 𝑦-values, which we can do by substituting each 𝑥-value in turn
into the linear equation which, remember, that was 𝑦 is equal to 11 minus
𝑥. When 𝑥 is equal to seven, we
find that 𝑦 is equal to 11 minus seven which is four. And when 𝑥 is equal to four,
we find that 𝑦 is equal to 11 minus four, which is seven. So we have exactly the same
values for 𝑦, as we did for 𝑥.
The reason for this is because
at the beginning of the problem, we just let the two numbers be represented by
𝑥 and 𝑦. We didn’t specify whether 𝑥 or
𝑦 was the larger number. So we found the same solution
twice. The two numbers are seven and
four. And either can be 𝑥 and then
the other will be 𝑦. 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 — seven
squared plus four squared, which is 49 plus 16 — 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 we’re looking for are seven and four.
