Question

If alpha and beta are the roots of the equation x^2+px+q=0 then the equation whose roots are (alpha/beta) and (beta/alpha) are

Answer 1

Answer:

Step-by-step explanation:

Given, α and β are the roots of the equation x²+px+q=0

So,

$\sf{\blue{Sum\,\,of\,\,roots,\,(\alpha+\beta)=-\dfrac{p}{1}=-p}}$

$\sf{\green{Products\,\,of\,\,roots,\,(\alpha\beta)=\dfrac{q}{1}=q}}$

Now,

$\rm{\bold{\purple{Equation\,\,\,whose\,\,\,roots\,\,\,are\,\,\,\dfrac{\alpha}{\beta}\,\,\,and\,\,\,\dfrac{\beta}{\alpha}\,\,\,is}}}$

$\rm{k\bigg[x^2+\bigg(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\bigg)x+\bigg(\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha}\bigg)\bigg]}$

Where 'k' is any real number

$\rm{=k\bigg[x^2+\bigg(\dfrac{\alpha^2+\beta^2}{\alpha\beta}\bigg)x+1\bigg]}$

$\rm{=k\bigg[x^2+\bigg\{\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\bigg\}x+1\bigg]}$

Now, put the values of (α+β) and (αβ),

$\rm{=k\bigg[x^2+\bigg\{\dfrac{(-p)^2-2(q)}{q}\bigg\}x+1\bigg]}$

$\rm{=k\bigg[x^2+\bigg\{\dfrac{p^2-2q}{q}\bigg\}x+1\bigg]}$

$\rm{=\dfrac{k}{q}\bigg[q\,x^2+(p^2-2q)x+q\bigg]}$

Put  $\sf{n=\dfrac{k}{q}}$, where n is any real number

$\rm{=n[q\,x^2+(p^2-2q)x+q]}$