Question Video: Forming Quadratic Equations in the Simplest Form Using the Relation between Their Coefficients and Roots Mathematics

Given that 𝐿 and 𝑀 are the roots of the equation π‘₯Β² βˆ’ 13π‘₯ βˆ’ 5 = 0, find, in its simplest form, the quadratic equation whose roots are 𝐿 + 1 and 𝑀 + 1.

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Video Transcript

Given that 𝐿 and 𝑀 are the roots of the equation π‘₯ squared minus 13π‘₯ minus five equals zero, find, in its simplest form, the quadratic equation whose roots are 𝐿 plus one and 𝑀 plus one.

Let’s begin by recalling the relationship between a quadratic equation in this form, one where the coefficient of π‘₯ squared is equal to one, and the roots of that equation. When we have an equation in this form, that is, we have a quadratic expression equal to zero where the coefficient of the π‘₯ squared term is one, we can say that the negative coefficient of the π‘₯-term is equal to the sum of the roots of the equation and the constant term is equal to the product of the roots.

Now, we’re told that 𝐿 and 𝑀 are the roots of our equation π‘₯ squared minus 13π‘₯ minus five equals zero. The sum of these roots must be the sum of 𝐿 and 𝑀; that’s 𝐿 plus 𝑀. Now that’s equal to the negative coefficient of π‘₯. Well, the coefficient of π‘₯ here is negative 13. And so, 𝐿 plus 𝑀 must be equal to positive 13. Then, the product of these roots is 𝐿 times 𝑀. Now that’s just equal to the constant term. So, it’s equal to negative five.

And so, we have a pair of equations in 𝐿 and 𝑀. The first is 𝐿 plus 𝑀 equals 13 and the second is 𝐿 times 𝑀 equals negative five. We could look to solve these equations simultaneously; however, we’re looking to find a quadratic equation whose roots are 𝐿 plus one and 𝑀 plus one. And so, if we can find the sum of 𝐿 plus one and 𝑀 plus one and the value of the product of these two expressions, we’ll be able to go straight into writing that quadratic equation.

Let’s look at our first equation, 𝐿 plus 𝑀 equals 13. And we’re going to manipulate this equation, so we include the new roots. Those are 𝐿 plus one plus 𝑀 plus one. But that, of course, is simply equal to 𝐿 plus 𝑀 plus two. Now, if 𝐿 plus 𝑀 is equal to 13, 𝐿 plus 𝑀 plus two must be equal to 13 plus two, and that’s equal to 15. So the sum of the roots of our new equation is 15. And we’ll be able to substitute that into the general form in a moment.

But what are we going to do about the product of these roots? Well, the product of the new roots is 𝐿 plus one times 𝑀 plus one. Let’s distribute these parentheses in the usual way. We multiply the first term in each expression to get 𝐿𝑀. We then multiply the outer terms, 𝐿 times one is 𝐿, and then the inner terms, one times 𝑀 is 𝑀. Finally, we multiply the last term in each expression. Well, one times one is equal to one. And so, the product of our new roots is 𝐿𝑀 plus 𝐿 plus 𝑀 plus one.

But remember, we know 𝐿𝑀 is equal to negative five; 𝐿 plus 𝑀 is equal to 13. And so, this whole expression becomes negative five plus 13 plus one, which is equal to nine. So, for our new quadratic equation, the sum of the roots is 15 and their product is nine. And so, substituting all of this into our original formula, we find its equation is as shown. The equation we’re interested in is π‘₯ squared minus 15π‘₯ plus nine equals zero.

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