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) ?

3 Answers
Feb 22, 2018

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

Feb 22, 2018

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).

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