If 
are the zeros of the quadratic polynomial 
,from a polynomial whose zeros are 
Answers
SOLUTION :
Given : α and β are the zeroes of the quadratic polynomial f(x) = x² + px + q
On comparing with ax² + bx + c,
a = 1 , b = p , c = q
Sum of the zeroes = −coefficient of x / coefficient of x²
α + β = -b/a = - p/1 = -p
α + β = - p……………………….(1)
Product of the zeroes = constant term/ Coefficient of x²
αβ = c/a = q/1 = q
αβ = q ………………………..(2)
ATQ
Sum of the zeroes of required polynomial = (α + β)² + (α–β)²
= (α + β)² + (α² + β² –2αβ)
[(a - b )² = a² + b² -2ab]
= (α+β)² +(α + β)² –2αβ–2αβ
[From this identity (a + b )² = a² + b² + 2ab , we get (a + b )² - 2ab = a² + b² ]
= (α + β)² +(α + β)² –4αβ
= (−p)² + (−p)² - 4×q
[From eq 1 & 2]
= p² + p² –4q
= 2p² –4q ………………..(3)
Product of the zeroes of required polynomial = (α + β)² (α–β)²
(α + β)² (α² + β² –2αβ)
(α + β)² (α+β)² –2αβ–2αβ
(α + β)² (α+β)² –4αβ
= (- p)² (- p)² –4q)
[From eq 1 & 2]
= p² (p² –4q)................(4)
So, the quadratic polynomial is ,
k(x²–(sum of the zeroes)x + (product of the zeroes)
[ k is any non zero real integer]
= k(x² –(2p² - 4q)x + p² (p² - 4q)
[From eq 3 & 4]
Hence, the required quadratic polynomial is f(x)= k( x² –(2p² - 4q)x + p² (p² - 4q)
HOPE THIS ANSWER WILL HELP YOU...
Answer:
f(x)=x
2
+px+q
Roots are α and β
α+β=−p
αβ=q
(α+β)
2
=p
2
(α−β)
2
=(α+β)
2
−4αβ
p
2
−4q
New Roots are (α+β)
2
and (α−β)
2
Sum=(α+β)
2
+ (α−β)
2
=p
2
+p
2
−4q
=2(p
2
−2q)
Product=(α+β)
2
(α−β)
2
=p
2
(p
2
−4q)
∴f(x)=K(x
2
−2(p
2
−2q)x+p
2
(p
2
−4q))