### Video Transcript

Given that πΏ and π are the roots of the equation two π₯ squared minus 10π₯ plus one equals zero, find, in its simplest form, the quadratic equation whose roots are πΏ over three and π over three.

We begin by recalling some key facts about a quadratic equation written in the form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are constants and π is nonzero. If the two roots of the quadratic are π sub one and π sub two, their sum is equal to negative π over π. The product of the two roots, π one multiplied by π two, is equal to π over π.

In this question, we are given the equation two π₯ squared minus 10π₯ plus one equals zero. The values of π, π, and π here are two, negative 10, and one. We are also told that πΏ and π are the two roots of the equation. Therefore, the sum of these, πΏ plus π, is equal to negative negative 10 over two. This is equal to five. The product of the roots πΏ multiplied by π is equal to one over two or one-half. We can now use these values to help find the quadratic equation whose roots are πΏ over three and π over three.

Letβs begin by considering the sum of these roots. We have πΏ over three plus π over three. As the denominators are the same, we can simply add the numerators, giving us πΏ plus π over three. We have already worked out that πΏ plus π is equal to five. This means that the sum of the roots is equal to five over three or five-thirds. Negative π over π is equal to five-thirds.

We will now consider the product of our two roots. We need to multiply πΏ over three and π over three. We do this by multiplying the numerators and denominators separately, giving us πΏπ over nine. As πΏπ is equal to one-half, we have a half over nine or a half divided by nine. This is equal to one eighteenth. The product of our roots π over π is equal to one eighteenth.

We can now solve these two equations to find the values of π, π, and π. We note that the denominators on the right-hand side are different. However, on the left-hand side of both our equations, the denominator is π. We can therefore multiply the numerator and denominator of the right-hand side of our first equation by six. Five over three or five-thirds is the same as 30 over 18.

Our next step is to let π equal 18. From the first equation, this means that negative π is equal to 30, which means that π is equal to negative 30. From the second equation, if π over π is equal to one over 18 and π equals 18, then π must be equal to one. The quadratic equation whose roots are πΏ over three and π over three is 18π₯ squared minus 30π₯ plus one equals zero.