If 𝐿 and 𝑀 are the roots of the equation 𝑥 squared plus 17𝑥 plus one equals zero, find, in its simplest form, the quadratic equation whose roots are three 𝐿 and three 𝑀.
Let’s begin by recalling some information about the quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero. If the roots of this equation are 𝑟 sub one and 𝑟 sub two, the sum of these roots is equal to negative 𝑏 over 𝑎. The product of the roots 𝑟 one multiplied by 𝑟 two is equal to 𝑐 over 𝑎.
We are given the equation 𝑥 squared plus 17𝑥 plus one is equal to zero. This means that the values of 𝑎, 𝑏, and 𝑐 are one, 17, and one, respectively. We are also told that the roots of this equation are 𝐿 and 𝑀. The sum of these roots 𝐿 plus 𝑀 is therefore equal to negative 17 over one. This is equal to negative 17. The product of the roots 𝐿 multiplied by 𝑀 is equal to one over one. This is equal to one. We now need to find the quadratic equation whose roots are three 𝐿 and three 𝑀.
Let’s begin by considering the sum of these roots. This is equal to three 𝐿 plus three 𝑀. And these terms have a common factor of three. We can, therefore, rewrite this as three multiplied by 𝐿 plus 𝑀. As 𝐿 plus 𝑀 is equal to negative 17, we need to multiply three by negative 17. This is equal to negative 51. In our new quadratic equation, negative 𝑏 over 𝑎 is equal to negative 51.
Let’s now consider the product of our roots. We need to multiply three 𝐿 by three 𝑀. This simplifies to nine 𝐿𝑀. And as 𝐿𝑀 is equal to one, we have nine multiplied by one. 𝑐 over 𝑎 is, therefore, equal to nine.
We now have two equations that we can solve to calculate the values of 𝑎, 𝑏, and 𝑐. As negative 51 and nine are integers, we can let 𝑎 equal one. From the first equation, this means that negative 𝑏 is equal to negative 51, which in turn means that 𝑏 is equal to 51. From the second equation, if 𝑎 is equal to one, 𝑐 is equal to nine.
The quadratic equation whose roots are three 𝐿 and three 𝑀 is 𝑥 squared plus 51𝑥 plus nine equals zero.