If alpha and beta are the roots of px^2+qx+r=0 Form the equation whose roots are (2alpha+1/alpha), (2beta+1/beta)?

2 Answers
Narad T.
May 1, 2017

The equation is #prx^2+q(2r+p)x+(4r^2+p^2+2q^2)-4pr=0#

Explanation:

The equation is

#px^2+qx+r=0#

#x^2+q/px+r/p=0#

If #alpha# and #beta# are the roots, then

the sum of the roots is #alpha+beta=-q/p#

and

the product of the roots is #alpha beta=r/p#

#=alpha^2+beta^2=(alpha+beta)^2-2alphabeta=q^2/p^2-2r/p#

The new roots are

#(2alpha+1/alpha)# and #(2beta+1/beta)#

Therefore, the sum of the roots is

#(2alpha+1/alpha)+(2beta+1/beta)#

#=2(alpha+beta)+(1/alpha+1/beta)#

#=2(alpha+beta)+(alpha+beta)/(alpha beta)#

#=(alpha+ beta)(2+1/(alpha beta))#

#=-q/p*(2+1/(r/p))#

#=-q/p*(2r+p)/r#

and

The product of the roots is

#(2alpha+1/alpha)(2beta+1/beta)#

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

#=4alpha beta+2(alpha^2+beta^2)/(alpha beta)+1/(alpha beta)#

#=4r/p+2*(q^2/p^2-2r/p)/(r/p)+1/(r/p)#

#=4r/p+p/r+2p/r*(q^2/p^2-2r/p)#

#=4r/p+p/r+2q^2/(rp)-4#

#=(4r^2+p^2+2q^2)/(rp)-4#

The new equation is

#x^2+(q/p*(2r+p)/r)x+(4r^2+p^2+2q^2)/(rp)-4=0#

#prx^2+q(2r+p)x+(4r^2+p^2+2q^2)-4pr=0#

Shwetank Mauria
May 1, 2017

#rpx^2+(2qr+pq)x+(4r^2+p^2+2q^2-4rp)=0#

Explanation:

If roots of an equation are #alpha# and #beta#,, the equation is

#(x-alpha)(x-beta)=0# i.e. #x^2-(alpha+beta)+alphabeta=0#

The equation #px^2+qx+r=0# can be written as

#x^2+q/px+r/p=0#. Hence, we have

sum of the roots #alpha+beta=-q/p# and product of roots #alphabeta=r/p#

Equation with roots as #2alpha+1/alpha# and #2beta+1/beta# is therefore

#x^2-(2alpha+1/alpha+2beta+1/beta)x+(2alpha+1/alpha)(2beta+1/beta)=0#

Let us now workout #2alpha+1/alpha+2beta+1/beta# and #(2alpha+1/alpha)(2beta+1/beta)#

#2alpha+1/alpha+2beta+1/beta#

= #2(alpha+beta)+(alpha+beta)/(alphabeta)#

= #2xx(-q/p)+(-q/p)/(r/p)=-2q/p-q/r=(-2qr-pq)/(rp)#

and #(2alpha+1/alpha)(2beta+1/beta)#

= #4alphabeta+1/(alphabeta)+2(alpha/beta+beta/alpha)#

= #4r/p+1/(r/p)+2(alpha^2+beta^2)/(alphabeta)#

= #4r/p+p/r+2((alpha+beta)^2-2alphabeta)/(alphabeta)#

= #4r/p+p/r+2((-q/p)^2-2r/p)/(r/p)#

= #4r/p+p/r+2q^2/p^2xxp/r-4#

= #4r/p+p/r+2q^2/(rp)-4#

= #(4r^2+p^2+2q^2-4rp)/(rp)#

and hence equation is

#x^2-((-2qr-pq)/(rp))x+(4r^2+p^2+2q^2-4rp)/(rp)=0#

or #rpx^2+(2qr+pq)x+(4r^2+p^2+2q^2-4rp)=0#

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