Question

If alpha and beta are the zeros of x^2+pc+q, find (Alpha/beta +2)(beta/alpha+2) ?

If `alpha and beta` are the zeros of `x^2+px+q,` find
`(alpha/beta +2)(beta/alpha+2) ?`

Answer 1

The answer is `=(2p^2+q)/q`

Explanation:

The quadratic equation is

`x^2+px+q=0`

The roots are `alpha` and `beta`

Therefore,

`alpha+beta=-p`... .....`(1)`

and

`alphabeta=q`..........`(2)`

We need

`(alpha/beta+2)(beta/alpha+2)=1+2alpha/beta+2beta/alpha+4`

`(alpha/beta+beta/alpha=(alpha^2+beta^2)/(alphabeta))`

`=((alpha+beta)^2-2alphabeta)/(q)`

`=(p^2-2q)/q`

Finally,

`(alpha/beta+2)(beta/alpha+2)=5+2*(p^2-2q)/q`

`=(5q+2p^2-4q)/(q)`

`=(2p^2+q)/q`

Answer 2

`2*p^2/q+1, or, 1/q(2p^2+q)`.

Explanation:

Given that `alpha and beta` are the zeroes of `x^2+px+q`, we have

`alpha+beta=-p, and, alpha*beta=q.................(ast)`.

`"Now, the reqd. value"=(alpha/beta+2)(beta/alpha+2)`,

`=4+2(alpha/beta+beta/alpha)+(alpha/beta)(beta/alpha)`,

`=4+{2(alpha^2+beta^2)}/(alpha*beta)+1`,

`={4alpha*beta+2(alpha^2+beta^2)}/(alpha*beta)+1`,

`={2(alpha+beta)^2}/(alpha*beta)+1`,

`={2(-p)^2}/q+1................[because, (ast)]`,

`=2*p^2/q+1, or, 1/q(2p^2+q)`.

Answer 3

Thus,
`(alpha/beta+2)(beta/alpha+2)=(q+2p^2)/q`

Explanation:

Given:
`alpha`is a root of `x^2+px+q=0`
`beta`is a root of `x^2+px+q=0`
`"If ax^2+bx+c=0, " then'`
`alpha + beta = -b/a`
Here, `a=1 b=p, c=q,`
Substituting
`alpha + beta =-p/1=-p`
`alphabeta = c/a=q/1=q`
`(alpha/beta+2)(beta/alpha+2)=(alpha+2beta)/betaxx(beta+2alpha)/alpha`
`=((alpha+2beta)(beta+2alpha))/(alphabeta)`

`=(alphabeta+2beta^2+2alpha^2+4alphabeta)/(alphabeta)`

`=(alphabeta+2(alpha+beta)^2)/(alphabeta)`

`=(q+2(-p)^2)/q`

`(alpha/beta+2)(beta/alpha+2)=(q+2p^2)/q`

Thus,
`(alpha/beta+2)(beta/alpha+2)=(q+2p^2)/q`