Question Video: Finding an Unknown in a Quadratic Equation Using the Relation between Its Coefficient and Its Roots Mathematics

The roots of the equation π‘šπ‘₯Β² βˆ’ 12𝑛π‘₯ + 𝑙 = 0, where π‘š β‰  0, are 𝐿 and 𝑀. Given that 𝐿 > 𝑀 and 𝐿 βˆ’ 𝑀 = 20, does 𝐿 = 10 + (6𝑛/π‘š)?

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Video Transcript

The roots of the equation π‘šπ‘₯ squared minus 12𝑛π‘₯ plus 𝑙 equals zero, where π‘š is not equal to zero, are 𝐿 and 𝑀. Given that 𝐿 is greater than 𝑀 and 𝐿 minus 𝑀 equals 20, does 𝐿 equal 10 plus six 𝑛 over π‘š?

Okay, so in this question, what we have is a quadratic equation in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero. And what we have with a quadratic in this form is a couple of relationships to deal with the roots. And that is, first of all, that the sum of the roots is equal to negative 𝑏 over π‘Ž and the product of the roots is equal to 𝑐 over π‘Ž. Well, the first thing we’re gonna do before we use either of our relationships is take a look at 𝐿 minus 𝑀 equals 20.

Well, the first thing we want to do is rearrange our equation so that either 𝐿 or 𝑀 are the subject. So what we’re going to do is rearrange it so that 𝑀 is the subject because what we want to do eventually is get everything in terms of 𝐿. So, to enable us to do that, what we’re gonna do is add 𝑀 and subtract 20 from each side of the equation. So when we do that, we get 𝐿 minus 20 equals 𝑀. So, great, we’ve made 𝑀 the subject. So now, what’s the next step?

Well, now what we can do is use our first relationship and that is that the sum of the roots is equal to negative 𝑏 over π‘Ž. But to enable us to use this, the first thing we need to do is work out what the π‘Ž, 𝑏, and 𝑐 are from our quadratic. Well, our π‘Ž, 𝑏, and 𝑐 are π‘š, negative 12𝑛, and 𝑙. Well, if we substitute these values into our relationship for the sum of the roots, we can say that 𝐿 plus 𝑀, so the sum of our roots, is equal to 12𝑛 over π‘š. We get 12𝑛 as the numerator because it’s negative 𝑏 over π‘Ž and we had 𝑏 being equal to negative 12𝑛. So, negative negative 12𝑛 gives us positive 12𝑛.

So now, what we’re gonna do is substitute 𝑀 in for our 𝑀 because we know that 𝑀 is equal to 𝐿 minus 20. So when we do this, what we’re gonna get is 𝐿 plus 𝐿 minus 20 equals 12𝑛 over π‘š, which is gonna give us two 𝐿 minus 20 equals 12𝑛 over π‘š. Well, then, if we add 20 to both sides, what we’re gonna get is two 𝐿 equals 12𝑛 over π‘š plus 20. Well, great. Is this what we’re looking for? Is this the answer we’re looking to match? Well, no, because if we look at what we’ve got, we’re looking for 𝐿 equals, but here we got two 𝐿 equals.

So we do need another step. And that step is to divide through by two. And if we divide each term by two, we’re gonna get 𝐿 equals 12𝑛 over two π‘š plus 10. So if we take a look at the term which is a fraction, we can in fact divide through by two because two is a factor of 12 and two. So when we do this, what we’re gonna be left with is 𝐿 equals six 𝑛 over π‘š plus 10. Well, this is exactly what we’re looking for. So, therefore, we can say that yes, 𝐿 does equal 10 plus six 𝑛 over π‘š.

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