Question

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.

Answer 1

hope this helps ... this was an easy question... which class do you study?

Answer 2

Answer:

We have polynomial  f ( x ) =  x2  + p x  + q

And their roots are α  and  β

We know the relationship between zeros and coefficient . 

Sum of zeros  = −Coefficient of x / Coefficient of x²

which is -b/a.

So,

α  + β  =  - p                                         ------ ( 1 )

If we square on both 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 α β =  p²

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

And,

Products of zeros  =  Constant term / Coefficient of x²

So,

α  β  =  q   , if we 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, if we add equation 2 and 4, we get:

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

And we multiply equation 2 and 4, we get:

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

We know formula for polynomial, as the sum of zeros and product of zeros:

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 p2² - 4 q) x  + ( 2 p⁴ - 4 p⁴q) ] 

                             

= x²  - ( 2 p² - 4 q) x  + ( 2 p² - 4 p²q)  [ Here, the value of k is 1 ] 

HOPE THIS HELPS !!!