Question

step wise answer plz

plz answer soon for 50 points

i will mark u as brainlist​

Answer 1

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 will help you

Answer 2

Step-by-step explanation:

Given Quadratic Equation is f(x) = x² + px + q.

Here, a = 1, b = p, c = q.

α,β are the roots of the equation.

(i) Sum of roots:

α + β = -b/a

α + β = -p.

(ii) Product of roots:

αβ = c/a

αβ = q

Now,

∴ (α - β)² = (α + β)² - 4αβ

              = (-p)² - 4q

               = p² - 4q

Given Zeroes are (α + β)² and (α - β)².

f(x) = x² - (Sum of zeroes)x + (product of zeroes)

     = x² - {(α + β)² + (α - β)²} + (α + β)² * (α - β)²

     = x² - (p² + p² - 4q)x + p² * (p² - 4q)

     = x² - (2p² - 4q)x + p⁴ - 4qp²

Hope it helps!