Video: Forming Quadratic Equations in the Simplest Form given Their Roots

Given that 𝐿 and 𝑀 are the roots of the equation π‘₯Β² βˆ’ 2π‘₯ + 20 = 0, find, in its simplest form, the quadratic equation whose roots are 2 and 𝐿² + 𝑀².

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Video Transcript

Given that 𝐿 and 𝑀 are the roots of the equation π‘₯ squared minus two π‘₯ plus 20 equals zero, find, in its simplest form, the quadratic equation whose roots are two and 𝐿 squared plus 𝑀 squared.

We recall that any quadratic equation of the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero β€” which has two roots, π‘Ÿ sub one and π‘Ÿ sub two β€” then the sum of the roots is equal to negative 𝑏 over π‘Ž and the product of the two roots is equal to 𝑐 over π‘Ž. We are given the quadratic equation π‘₯ squared minus two π‘₯ plus 20 equals zero. Therefore, π‘Ž is equal to one, 𝑏 is equal to negative two, and 𝑐 is equal to 20. The roots of this equation are 𝐿 and 𝑀. Therefore, 𝐿 plus 𝑀 is equal to negative negative two over one. This is equal to two. The product of the two roots, 𝐿 multiplied by 𝑀, is equal to 20 over one. This is equal to 20.

Using this information, we need to find another quadratic equation whose roots are two and 𝐿 squared plus 𝑀 squared. In this equation, two plus 𝐿 squared plus 𝑀 squared must equal negative 𝑏 over π‘Ž and two multiplied by 𝐿 squared plus 𝑀 squared equals 𝑐 over π‘Ž, where π‘Ž, 𝑏, and 𝑐 are unknowns we need to calculate. We recall that expanding 𝐿 plus 𝑀 all squared gives us 𝐿 squared plus two 𝐿𝑀 plus 𝑀 squared. Subtracting two 𝐿𝑀 from both sides of this equation tells us that 𝐿 squared plus 𝑀 squared is equal to 𝐿 plus 𝑀 all squared minus two 𝐿𝑀. We notice that the expression on the right-hand side is contained in both of our equations. We also notice that 𝐿𝑀 is equal to 20 and 𝐿 plus 𝑀 is equal to two.

Substituting in these values, 𝐿 squared plus 𝑀 squared is equal to two squared minus two multiplied by 20. The left-hand side simplifies to negative 36. We can now substitute this value into both of our equations. Negative 𝑏 over π‘Ž is equal to two plus negative 36. This is equal to negative 34. Two multiplied by negative 36 is equal to 𝑐 over π‘Ž. 𝑐 over π‘Ž is therefore equal to negative 72. As both negative 34 and negative 72 are integers, we can let π‘Ž equal one. This means that negative 𝑏 is equal to negative 34, so 𝑏 equals 34. 𝑐 is equal to negative 72. The quadratic equation whose roots are two and 𝐿 squared plus 𝑀 squared is π‘₯ squared plus 34π‘₯ minus 72 is equal to zero.

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