Video Transcript
Given that 𝐿 and 𝑀 are the roots of the equation three 𝑥 squared minus six 𝑥 plus seven equals zero, find, in its simplest form, the quadratic equation whose roots are 𝐿 plus 𝑀 and 𝐿 times 𝑀.
Let’s begin by recalling what we know about the relationship between a quadratic equation and the roots of that equation. Let’s take a quadratic equation where the coefficient of 𝑥 squared is equal to one. In this case, we can say that the negative coefficient of 𝑥 in this equation gives us the sum of the roots of the equation. And the constant term tells us the product of those roots. Now, we might actually notice that our equation is not in this form. We have three 𝑥 squared minus six 𝑥 plus seven equals zero. However, we can manipulate it a little. We’re going to divide every single term in this equation by three. Now, of course, we can do that since zero divided by three is still zero. Three 𝑥 squared divided by three is 𝑥 squared. Then negative six 𝑥 divided by three is negative two 𝑥, and we’ll leave seven over three in fractional form.
So we have a quadratic equation in the right form now. It’s 𝑥 squared minus two 𝑥 plus seven over three. Let’s compare this equation to the general form. We said the negative coefficient of 𝑥 is the sum of our roots. Now our roots are 𝐿 plus 𝑀, and the coefficient of 𝑥 in our equation is negative two. The sum of our roots then must be negative negative two, which is simply two, so 𝐿 plus 𝑀 is equal to two. Then, the product of our roots, that’s 𝐿 times 𝑀, is equal to the constant term, so 𝐿𝑀 equals seven over three. And what we might look to do next is to solve these equations simultaneously. However, we’re told that we have a quadratic equation whose roots are 𝐿 plus 𝑀 and 𝐿𝑀. And so we’re going to use the equations we have to find the sum of these roots and the value for the product.
Let’s begin with the sum of our roots. The new roots are 𝐿 plus 𝑀 and 𝐿𝑀. So the sum of these roots is 𝐿 plus 𝑀 plus 𝐿𝑀. Now, in fact, if we look at this, we see it’s made up of two expressions for which we know the value. We know that 𝐿 plus 𝑀 is equal to two and 𝐿𝑀 is equal to seven-thirds. Now, in fact, let’s write two as six-thirds because we can then add these fractions. Six plus seven is 13, so six-thirds plus seven-thirds is thirteen-thirds, and that’s great. We have the sum of our new roots, and so we can find the coefficient of 𝑥 in our new equation. It will be negative 13 over three.
Let’s repeat this process and find the product of our new roots. It’s the product of 𝐿 plus 𝑀 and 𝐿𝑀, so it’s 𝐿𝑀 times 𝐿 plus 𝑀. Once again, we have the value of 𝐿𝑀 and the value of 𝐿 plus 𝑀. So the product of these is seven over three times two, which is simply 14 over three. Now that we know the product of the roots, we have the constant term for our new equation. We can therefore say that our equation must be 𝑥 squared minus 13 over three 𝑥 plus 14 over three equals zero. We’re going to write this in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. And we’ll do so by multiplying every single term by three. And so we have our new quadratic equation. It’s three 𝑥 squared minus 13𝑥 plus 14 equals zero.
