Video Transcript
Given that 𝐿 and 𝑀 are the roots of the equation 𝑥 squared minus six 𝑥 minus seven equals zero, find, in its simplest form, the quadratic equation whose roots are 𝐿 squared and 𝑀 squared.
One way of answering this question would be to begin by factoring our quadratic equation. We can factor the expression 𝑥 squared minus six 𝑥 minus seven into two sets of parentheses. The first term in each of these will be 𝑥, as 𝑥 multiplied by 𝑥 is 𝑥 squared. Our next step is to find two integers that have a sum of negative six and a product of negative seven. Negative seven multiplied by one is negative seven, and negative seven plus one is negative six. 𝑥 squared minus six 𝑥 minus seven can be rewritten as 𝑥 minus seven multiplied by 𝑥 plus one. As the product of our two parentheses equals zero, either 𝑥 minus seven equals zero or 𝑥 plus one equals zero. Solving these two equations gives us 𝑥 equals seven and 𝑥 equals negative one.
We are told that the two roots of the quadratic equation are 𝐿 and 𝑀. Therefore, we will let 𝐿 equal seven and 𝑀 equal negative one. 𝐿 squared is therefore equal to 49, as seven squared is 49. 𝑀 squared is equal to one. The roots of our new quadratic equation will be 49 and one. This means that our quadratic will have two factors: 𝑥 minus 49 and 𝑥 minus one. The product of these must equal zero. Distributing our parentheses using the FOIL method gives us 𝑥 squared minus 𝑥 minus 49𝑥 plus 49. And we know this is equal to zero. Collecting the like terms, we obtain the quadratic equation 𝑥 squared minus 50𝑥 plus 49 equals zero. This is the quadratic equation whose roots are 𝐿 squared and 𝑀 squared.
There is an alternative method we could’ve used here using our knowledge of roots of quadratics. We know that for any quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, the sum of the two roots will be equal to negative 𝑏 over 𝑎 and the product of the roots will be equal to 𝑐 over 𝑎. In our original equation, 𝑎 is equal to one, 𝑏 is equal to negative six, and 𝑐 is equal to negative seven. As the two roots are 𝐿 and 𝑀, 𝐿 plus 𝑀 is equal to negative negative six over one. This is equal to six. 𝐿 multiplied by 𝑀, the product of the roots, is equal to negative seven over one. This is equal to negative seven.
As our new quadratic has roots 𝐿 squared and 𝑀 squared, we need to calculate 𝐿 squared plus 𝑀 squared and 𝐿 squared multiplied by 𝑀 squared. 𝐿 squared plus 𝑀 squared is equal to 𝐿 plus 𝑀 all squared minus two 𝐿𝑀. We know this from our knowledge of the sum of two squares. As 𝐿 plus 𝑀 is equal to six and 𝐿𝑀 is equal to negative seven, we can substitute these into our expression. This gives us six squared minus two multiplied by negative seven. Six squared is equal to 36, and negative two multiplied by negative seven is 14. This means that 𝐿 squared plus 𝑀 squared is equal to 50.
𝐿 squared 𝑀 squared can be rewritten as 𝐿𝑀 all squared. This is therefore equal to negative seven squared, which in turn equals 49. In our new quadratic, the sum of our roots negative 𝑏 over 𝑎 is equal to 50 and the product of our roots 𝑐 over 𝑎 is equal to 49. As both of these values are integers, we can let 𝑎 equal one. This means that negative 𝑏 is equal to 50, and therefore 𝑏 is equal to negative 50. As 𝑐 over 𝑎 is equal to 49, 𝑐 is equal to 49. The quadratic equation written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero is therefore 𝑥 squared minus 50𝑥 plus 49 equals zero. This confirms the answer we obtained using our first method.
