Thread: Roots of the equation

1. Roots of the equation

Hello,

(1) Set $\displaystyle y=1/(px+q)$ to find the equation whose roots are $\displaystyle 1/(p\alpha+q)$ and $\displaystyle 1/(p\beta+q)$

Thank you.

2. Originally Posted by dynamicsagar
Hello,

(1) Set $\displaystyle y=1/(px+q)$ to find the equation whose roots are $\displaystyle 1/(p\alpha+q)$ and $\displaystyle 1/(p\beta+q)$

Thank you.
I don't see what the substitution is for. If we know that a polynomial has only two roots, $\displaystyle 1/(p\alpha+q)$ and $\displaystyle 1/(p\beta+q)$, each with multiplicity 1, then we can immediately write

$\displaystyle f(x)=k\left(x-\frac{1}{p\alpha+q}\right)\left(x-\frac{1}{p\beta+q}\right), k \in \mathbb{R}, k \ne 0$

Maybe the question wants it in this form?

$\displaystyle f(x)=k(x-y_{\alpha})(x-y_{\beta}), k \in \mathbb{R}, k \ne 0$

3. The remainder of the question

Hello,

Sorry; I forgot to type the first half of the question.

(1) Let $\displaystyle \alpha$ and $\displaystyle \beta$ be the roots of the equation $\displaystyle ax^2+bx+c=0$. The rest of the question goes as written in the first post.

Sagar

4. Originally Posted by dynamicsagar
Hello,
(1) Set $\displaystyle y=1/(px+q)$ to find the equation whose roots are $\displaystyle 1/(p\alpha+q)$ and $\displaystyle 1/(p\beta+q)$
There's a straightforward albeit messy way to express a new polynomial with those roots in terms of just a,b,c. The function I wrote above in terms of alpha and beta is symmetric in alpha and beta. Without loss of generality, let $\displaystyle \alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and let $\displaystyle \beta=\frac{-b-\sqrt{b^2-4ac}}{2a}$. Substitute in and you have your equation. (Choice of $\displaystyle k$ is arbitrary; for simplicity, you can let $\displaystyle k=1$.)
I still don't know what setting $\displaystyle y=1/(px+q)$ is supposed to accomplish, but maybe I could figure it out by looking at an example in your book. Possibly it's something obvious that I'm just not seeing. If you have the intended solution and wish to post it, I'd be curious to see.