Math, asked by Darkhawk, 1 year ago

If alpha and beta are the zeroes of the quadratic polynomial f(x) = x2 - px + q, prove that alpha2/beta2 + beta2/alpha2= p4/q2 - 4p2/q + 2?

Answers

Answered by ahmed8486
310

We have polynomial  f ( x ) =  x2  + p x  + q
And
Roots are α  and β
And
we know from relationship between zeros and coefficient . 

Sum of zeros  = −Coefficient of xCoefficient of x2
So,
α  + β  =  - p                                         ------ ( 1 )

Taking whole square on both hand side , we get

( α  + β  )2 = p 2                                  ------ ( 2 )

⇒α2 + β2 +  2 α β = p2⇒α2 + β2 +  2 α β − 2αβ + 2αβ=  p2⇒α2 + β2 −  2 α β  +4 αβ=  p2⇒(α  − β) 2  +4 αβ=  p2                     −−−− ( 3 )

And

Products of zeros  = Constant termCoefficient of x2
So,
α  β  =  q        , Substitute that value in equation 3 , we get

⇒(α− β)2 +  4 (q ) = p2⇒(α− β)2 +4 q=  p2⇒(α− β)2  = p2 − 4 q              −−−− ( 4 )   

Now we add equation 2 and 4 and get

(α + β)2 + (α − β)2 = p2 +  p2 − 4 q= 2 p2 − 4 q

And we multiply equation 2 and 4 and get

(α + β)2 × (α − β)2 = p2( p2 − 4 q)= p4 − 4 p2q

And we know formula for polynomial when sum of zeros and product of zeros we know :

Polynomial  =  k [ x2  - ( Sum of zeros ) x  + ( Product of zeros ) ]   , Here k is any non zero real number.

Substitute values , we get

Quadratic polynomial  =  k [ x2  - ( 2 p2 - 4 q) x  + ( 2 p4 - 4 p2q) ] 

                             
= x2  - ( 2 p2 - 4 q) x  + ( 2 p4 - 4 p2q) [ taking k = 1 ]                                         ( Ans )

Hope this information will clear your doubts about topic.
Answered by sam200431
432

here is the answer hope it help full for u

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