### 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.