Question

ifα ,β are roots of x^2-px+q=0 andα-2,β+2 are roots of x^2-px+r=0 then prove that 16q +(r+4-q)^2=4p^2

Answer 1

hope the answer helps

Answer 2

Step-by-step explanation:

Given that,

α,β are the roots of x

2

−px+q=0

Sum of roots =α+β=

a

−b

=p

Product of roots =αβ=

a

c

=q

Difference of roots =α−β=

a

b

2

−4ac

=

p

2

−4q

∵α−2,β+2 are the roots of x

2

−px+r=0

Sum of roots =α+β=

a

−b

=p

Product of roots =(α−2)(β+2)=

a

c

=r

⟹αβ+2(α−β)−4=r

⟹q+2(

p

2

−4q

)−4=r

⟹2(

p

2

−4q

)=r+4−q

Squaring on both sides, we get

4(p

2

−4q)=(r+4−q)

2

⟹4p

2

−16q=(r+4−q)

2

∴4p

2

=16q+(r+4−q)

2

Hence proved.