Video Transcript
If 𝐿 and 𝑀 are the roots of the equation two 𝑥 squared minus three 𝑥 plus one equals zero, find, in its simplest form, the quadratic equation whose roots are two 𝐿 squared and two 𝑀 squared.
We might recall a relationship between a quadratic equation whose coefficient of 𝑥 squared is one and the roots of that equation. The negative coefficient of 𝑥 tells us the sum of the roots of that equation, and then the constant term tells us the product. Now, of course, we’re told that 𝐿 and 𝑀 are the roots of our equation, but it’s not currently in this form. So we’re going to divide each term in this equation by two to achieve this. Two 𝑥 divided by two is 𝑥 squared. We then divide the coefficient of 𝑥 by two and we get negative three over two 𝑥. Finally, we divide the constant term by two. And of course, we know that zero divided by two is zero.
So we now have the equation in the correct form. And so we’re able to work out the value of the sum of our roots, that’s 𝐿 plus 𝑀, and their product, that’s 𝐿 times 𝑀. 𝐿 plus 𝑀, the sum of the roots, is the negative coefficient of 𝑥. Now, since the coefficient of 𝑥 here is negative three over two, we can say that the sum of our roots must simply be three over two. Then the product of the roots is equal to the constant term. Well, that’s quite simply one-half. So we have that 𝐿 plus 𝑀 is equal to three over two and then 𝐿𝑀 is equal to one-half.
Let’s now think about the roots of our new equation. They are still in terms of 𝐿 and 𝑀. They’re two 𝐿 squared and two 𝑀 squared. This time, if we find an expression for the sum of the roots and the product of the roots, we get two 𝐿 squared plus two 𝑀 squared and two 𝐿 squared times two 𝑀 squared. So we need to use our earlier expressions for 𝐿 plus 𝑀 and 𝐿𝑀 to find values for these expressions. Let’s look at the equation we formed for the sum of our roots. And we’re going to square it. On the left-hand side, we get 𝐿 plus 𝑀 squared, and then we can just simply square the numerator and the denominator of three over two, and we get nine over four. But of course, 𝐿 plus 𝑀 squared is 𝐿 plus 𝑀 multiplied by 𝐿 plus 𝑀. So let’s distribute these parentheses.
When we do, we get 𝐿 squared plus two 𝐿𝑀 plus 𝑀 squared equals nine over four. And we now notice that we have 𝐿 squared plus 𝑀 squared. So we’re on our way to find an expression for the sum of our new roots. What we’ll do next is we’ll replace 𝐿𝑀 with the value of one-half. And so we get 𝐿 squared plus two times a half plus 𝑀 squared equals nine over four. Now, of course, two times a half is one. So we get 𝐿 squared plus one plus 𝑀 squared equals nine over four. And then if we subtract one from both sides, we find that expression for 𝐿 squared plus 𝑀 squared. Now, in fact, we can write one as four over four. And the right-hand side becomes nine over four minus four over four, and that’s simply five over four.
Now we notice that we have an expression for 𝐿 squared plus 𝑀 squared. But we want an expression for two 𝐿 squared plus two 𝑀 squared. And so to achieve this, we’ll multiply the left-hand side by two, which means we also have to do the same to the right-hand side. The distributive property of multiplication tells us that this is equal to two 𝐿 squared plus two 𝑀 squared. Then five over four times two is five over two. And we can therefore say that the sum of the roots of our new quadratic equation must be five over two.
But what about the product of these roots? Let’s begin by multiplying two 𝐿 squared by two 𝑀 squared. And we get four 𝐿 squared 𝑀 squared. We’re going to clear some space and go back to one of our earlier equations. Let’s look at the equation 𝐿𝑀 equals one-half. It should be quite clear to us that if we square this, we can get the 𝐿 squared 𝑀 squared part. 𝐿𝑀 squared is 𝐿 squared 𝑀 squared. And then one-half squared is a quarter. Of course, we want four 𝐿𝑀 squared. So we’re going to multiply both sides of this equation by four. A quarter times four is simply equal to one. And so we can say that four 𝐿 squared 𝑀 squared, which is the product of our roots, is simply one.
We replace these values in the general form. We know that the sum of our roots is five over two and the product is one. And so we see that our new quadratic equation is 𝑥 squared minus five over two 𝑥 plus one equals zero. We’ll perform one more step, and we’ll make each term integer by multiplying through by two. And when we do, we see that the quadratic equation whose roots are two 𝐿 squared and two 𝑀 squared is two 𝑥 squared minus five 𝑥 plus two equals zero.
