### Video Transcript

If πΏ and π are the roots of the equation two π₯ squared minus three π₯ plus one equals zero, find, in its simplest form, the quadratic equation whose roots are two πΏ squared and two π squared.

We might recall a relationship between a quadratic equation whose coefficient of π₯ squared is one and the roots of that equation. The negative coefficient of π₯ tells us the sum of the roots of that equation, and then the constant term tells us the product. Now, of course, weβre told that πΏ and π are the roots of our equation, but itβs not currently in this form. So weβre going to divide each term in this equation by two to achieve this. Two π₯ divided by two is π₯ squared. We then divide the coefficient of π₯ by two and we get negative three over two π₯. 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 weβre able to work out the value of the sum of our roots, thatβs πΏ plus π, and their product, thatβs πΏ times π. πΏ plus π, the sum of the roots, is the negative coefficient of π₯. Now, since the coefficient of π₯ 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, thatβs quite simply one-half. So we have that πΏ plus π is equal to three over two and then πΏπ is equal to one-half.

Letβs now think about the roots of our new equation. They are still in terms of πΏ and π. Theyβre two πΏ squared and two π squared. This time, if we find an expression for the sum of the roots and the product of the roots, we get two πΏ squared plus two π squared and two πΏ squared times two π squared. So we need to use our earlier expressions for πΏ plus π and πΏπ to find values for these expressions. Letβs look at the equation we formed for the sum of our roots. And weβre going to square it. On the left-hand side, we get πΏ plus π 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, πΏ plus π squared is πΏ plus π multiplied by πΏ plus π. So letβs distribute these parentheses.

When we do, we get πΏ squared plus two πΏπ plus π squared equals nine over four. And we now notice that we have πΏ squared plus π squared. So weβre on our way to find an expression for the sum of our new roots. What weβll do next is weβll replace πΏπ with the value of one-half. And so we get πΏ squared plus two times a half plus π squared equals nine over four. Now, of course, two times a half is one. So we get πΏ squared plus one plus π squared equals nine over four. And then if we subtract one from both sides, we find that expression for πΏ squared plus π 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 thatβs simply five over four.

Now we notice that we have an expression for πΏ squared plus π squared. But we want an expression for two πΏ squared plus two π squared. And so to achieve this, weβll 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 πΏ squared plus two π 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? Letβs begin by multiplying two πΏ squared by two π squared. And we get four πΏ squared π squared. Weβre going to clear some space and go back to one of our earlier equations. Letβs look at the equation πΏπ equals one-half. It should be quite clear to us that if we square this, we can get the πΏ squared π squared part. πΏπ squared is πΏ squared π squared. And then one-half squared is a quarter. Of course, we want four πΏπ squared. So weβre 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 πΏ squared π 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 π₯ squared minus five over two π₯ plus one equals zero. Weβll perform one more step, and weβll make each term integer by multiplying through by two. And when we do, we see that the quadratic equation whose roots are two πΏ squared and two π squared is two π₯ squared minus five π₯ plus two equals zero.