Question

If alpha and beta are the zeros of the polynomial ax2+bx+c then form the polynomial whose zeroes are 1/alpha and 1/beta

Answer 1

x2-(alpha + beta)x+ alpha×beta.
x2-(1/alpha +1/beta)x+1/alpha×1/beta
x2-(alpha+beta/alpha× beta)x+ 1/alpha×beta
x2-( -ba/ca)x+a/c
ax2+(ba/ca)x+a/c
ax2+b/cx+a/c=0.

Answer 2

Answer:

$x^2-(\frac{bc-a^2}{ab})x-\frac{a^2}{ab}$

Step-by-step explanation:

Since, $\alpha$ and $\beta$ are the roots of the polynomial ax²+bx+c.

Thus, we can write,

$\alpha =-\frac{b}{a}\implies \frac{1}{\alpha}=-\frac{a}{b}$

And, $\beta = \frac{c}{a}\implies \frac{1}{\beta}=\frac{c}{a}$

Since, the equation that has the roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$is,

$x^2-(\frac{1}{\alpha}+\frac{1}{\beta})x+\frac{1}{\alpha}\frac{1}{\beta}$

$x^2-(-\frac{a}{b}+\frac{c}{a})x-\frac{a^2}{ba}$

$x^2-(\frac{-a^2+bc}{ab})x-\frac{a^2}{ab}$

$x^2-(\frac{bc-a^2}{ab})x-\frac{a^2}{ab}$

Which is the required equation.