Math, asked by nitii01, 9 months ago

if alpha and beta are the zeros of the quadratic polynomial f(x)= x^2-px+q , prove that α^2/β^2 +β^2/α^2 = p^4/q^2 - 4p^2/q +2​

Answers

Answered by harshu00705
4

Answer:

Step-by-step explanation:

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

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