Question

If α,β are the roots of x²+ax-b=0 and γ,δ are the roots of x²+ax+b=0 then (α-γ)(α-δ)(β-δ)(β-γ)=?

Answer 1

`= 64b^2`

Explanation:

`alpha = (- a + Lambda)/(2), qquad beta = (- a - Lambda)/(2), qquad Lambda = sqrt(a^2 + 4 b)`

`gamma = (- a + Gamma)/(2), qquad delta = (- a - Gamma)/(2), qquad Gamma = sqrt(a^2 - 4 b)`

• `alpha - gamma = Lambda - Gamma qquad qquad = P`

• `alpha - delta = Lambda + Gamma qquad qquad = Q`

• `beta - delta = - (Lambda - Gamma) qquad = R`

• `beta - gamma = - (Lambda + Gamma) qquad = S`

`PQRS = ((Lambda - Gamma)(Lambda + Gamma))^2`

`= (Lambda^2 - Gamma^2)^2`

`= ((a^2 + 4b) - (a^2 - 4b))^2`

`= (8b)^2 = 64b^2`

Answer 2

`(alpha-gamma)(alpha-delta)(beta-delta)(beta-gamma)=4(alphabeta)^2=4b^2`

Explanation:

We know that,

" If , `l and m` are the roots of `Ax^2+Bx+C=0` ,then

`(i)` the SUM of the Roots `=l+m=-B/A`

`(ii)` the PRODUCT of the Roots `=l*m=C/A`

...................................................................................................

Here,

`alpha and beta` are the roots of `x^2+ax-b=0`

So, `color(red)(alpha+beta=-a ,and alpha*beta=-bto(1)`

We have,

`gamma and delta` are the roots of `x^2+ax+b=0`

So, `color(red)(gamma+delta=-a , and gamma*delta=bto(2)`

From `(1) and (2)`

`color(blue)(gamma+delta=alpha+beta and gamma*delta=-alpha*betato(3)`

Let,

`K=(alpha-gamma)(alpha-delta)(beta-delta)(beta-gamma)`

`K=[alpha^2-alphagamma-alphadelta+gamma*delta][beta^2-gammabeta-deltabeta+gamma*delta]`

`K=[alpha^2-alphacolor(blue)((gamma+delta)+gamma*delta)][beta^2-betacolor(blue)((gamma+delta)+gamma*delta)]`

Using `(3)`we get

`K=[alpha^2-alphacolor(blue)((alpha+beta)-alpha*beta)][beta^2-betacolor(blue)((alpha+beta)-alphabeta)]`

`K=[alpha^2-alpha^2-alphabeta-alphabeta][beta^2-alphabeta-beta^2-alphabeta]`

`K=[-2alphabeta][-2alphabeta]`

`K=4(alphabeta)^2 tocolor(red)(Apply(1)`

`K=4(-b)^2=4b^2`

Hence,

`(alpha-gamma)(alpha-delta)(beta-delta)(beta- gamma)=4(alphabeta)^2=4b^2`