Question

The roots of the equation `x^2 + px + q =0`are β and α. The roots of another equation `x^2 + (q/2 +1)x + p -5=0` are α+2 and β+2, where α>β and p and q are constant. Find (a) the values of p and q ? (b) the value of α and β ?

(c) a quadratic equation with the given roots, 5α and 5β.

Answer 1

Given that roots of `x^2 + px + q =0` are `beta and alpha`.
So `alpha+beta =-p and alphabeta=q`

Again the roots of another equation `x^2 + (q/2 +1)x + p -5=0` are `beta +2and alpha+2`.

So `alpha+beta+4 =-(q/2+1) and (alpha+2)(beta+2)=p-5`

Now `alpha+beta+4 =-(q/2+1)`
`-p+4 =-(q/2+1)`

`=>2p-8 =q+2`

`=>2p-q=10........[1]`

Again

`(alpha+2)(beta+2)=p-5`

`=>alphabeta+2(alpha+beta)+4=p-5`

`=>q-2p+4=p-5`

`=>3p-q=9.......[2]`

Subtracting []1 from [2] we get
(a)
`p=-1`

So `q = -12`

Hence `alpha+beta=-p=1`
`alphabeta =q=-12`

`alpha -beta =sqrt((alpha+beta)^2-4alphabeta)=sqrt((-1)^2-4(-12))=7`
Given `alpha>beta`

(b)
So `2alpha=8=>alpha=4`
and `beta=-3`
(c) the equation having roots `5alpha and 5beta` will be

`x^2-(5alpha+5beta)x+25alphabeta=0`

`=>x^2-5(alpha+beta)x+25alphabeta=0`

`=>x^2-5x+25(-12)=0`

`=>x^2-5x-300=0`