Question Video: Forming a Quadratic Equation in the Simplest Form by Using the Relation between the Coefficient of a Quadratic Function and Its Roots Mathematics

Given that 𝐿 and π‘š are the roots of the equation π‘₯Β² βˆ’ 2π‘₯ + 5 = 0, find, in its simplest form, the quadratic equation whose roots are 𝐿² and π‘šΒ².

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Video Transcript

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

Let’s begin by recalling the relationship between a quadratic equation whose leading coefficient is one and its roots. We can represent it as π‘₯ squared minus the sum of the roots times π‘₯ plus the product of the roots equals zero. And so, this is essentially saying that if we have a quadratic equation equal to zero and the leading coefficient is one, in other words, the coefficient of π‘₯ squared is one, the negative coefficient of π‘₯ will tell us the sum of the roots and the constant term will tell us its product.

So we take the equation π‘₯ squared minus two π‘₯ plus five equals zero. The coefficient of π‘₯ is negative two. And so the sum must be the negative of negative two, so the sum of the roots must be positive two. Then the constant term is five. So the product of the roots must be five. So can we find two numbers that have a sum of two and a product of five? Well, no, not easily. We’re not going to get nice integer solutions. And so instead, we’re going to form equations using 𝐿 and π‘š. Since the sum of our roots is two and 𝐿 and π‘š are the roots, we can say 𝐿 plus π‘š must be equal to two. And then we can say that 𝐿 times π‘š is equal to five.

The roots of our new equation are 𝐿 squared and π‘š squared. And so, since their sum will be 𝐿 squared plus π‘š squared, we need to manipulate our equations to find an expression for 𝐿 squared plus π‘š squared and another expression for their product, 𝐿 squared π‘š squared. Let’s label our equations as one and two. We’re going to take the entirety of equation one and, we’re going to square it. In other words, we square both sides. So on the right-hand side, we get two squared, which is, of course, equal to four. Then, on the left-hand side, we get 𝐿 plus π‘š squared, which we can consider to be 𝐿 plus π‘š times 𝐿 plus π‘š.

And if we distribute these parentheses, we get 𝐿 squared plus two πΏπ‘š plus π‘š squared equals four. And then, if we subtract two πΏπ‘š from both sides, we get the expression for the sum of the roots 𝐿 squared and π‘š squared. It’s four minus two πΏπ‘š. But of course, we have an expression for πΏπ‘š; equation two tells us that πΏπ‘š is equal to five. And so, 𝐿 squared plus π‘š squared becomes four minus two times five, which is four minus 10 or simply negative six. So, we found the sum of the roots of our new equation and, therefore, the negative coefficient of π‘₯.

We’re now going to repeat this process for equation two; we’re going to square both sides. That is, πΏπ‘š squared equals five squared. But, of course, five squared is 25. And we can distribute to the power of two across both terms. And we get 𝐿 squared π‘š squared equals 25. We, therefore, find that the sum of our new roots is negative six and the product is 25. Let’s substitute these back into the general form. When we do, we get π‘₯ squared minus negative six π‘₯ plus 25 equals zero. And so, the quadratic equation whose roots are 𝐿 squared and π‘š squared is π‘₯ squared plus six π‘₯ plus 25 equals zero.

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