If 
are the zeros of the quadratic polynomial 
,find the polynomial whose roots are (i) 
(ii) 
Answers
SOLUTION :
Given : α and β are the zeroes of the quadratic polynomial f(x)= x² - 2x + 3
On comparing with ax² + bx + c,
a = 1 , b = -2 , c = 3
Sum of the zeroes = −coefficient of x / coefficient of x²
α + β = -b/a = - (-2)/1 = 2
α + β = 2 ……………………….(1)
Product of the zeroes = constant term/ Coefficient of x²
αβ = c/a = 3/1 = 3
αβ = 3 ………………………..(2)
(i) Sum of the zeroes of required polynomial = (α+2)+(β+2)
= α + β + 2 + 2
= (α + β) + 4
= 2 + 4 = 6
[From eq 1]
(α+2)+(β+2) = 6 …………………(3)
Product of the zeroes of required polynomial = (α + 2)(β + 2)
= β(α + 2) + 2 (α + 2)
=αβ + 2β + 2α + 4
= αβ + 2α + 2β + 4
= αβ + 2(α+β) + 4
= 3 + 2(2) + 4
[ From eq 1 & 2]
= 3 + 4 + 4
= 7 + 4 = 11
(α + 2)(β + 2) = 11………………(4)
So, quadratic polynomial is:
k(x²–(sum of the zeroes)x + (product of the zeroes)
[ k is any non zero real integer]
= k( x² - 6x + 11)
[ From eq 3 & 4]
Hence, the required quadratic polynomial is f(x)= k( x² - 6x + 11)
(ii) Sum of the zeroes of required polynomial = α–1/ α+1 + β–1/ β+1
= (α–1)(β+1)+(β–1)(α +1)/ (α+1)(β+1)
= αβ + α–β–1+ αβ + β–α–1/ (α+1)(β+1)
= αβ + αβ+ α – α - β + β –1 –1/ αβ + (α+β) + 1
= 2αβ -2/ αβ + (α+β) + 1
=2× 3 - 2 / 3 + (2) + 1
= 6 - 2 / 6
= 4/6 = ⅔
α–1/ α+1 + β–1/ β+1 = 2/3
Product of the zeroes of required polynomial = (α–1/ α+1) ×( β–1/ β+1)
= (α–1 × β–1) / (α + 1 × β + 1)
= αβ - β– α +1 / αβ + β+ α + 1
= αβ - ( β + α) +1 / αβ + (β+ α ) + 1
= 3 - 2 +1 / 3 + 2 +1
= 3 - 1 / 6
= 2 / 6 = ⅓
(α–1/ α+1) ×( β–1/ β+1) = 2/3
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 - 2x/3 + ⅓)
Hence, the required quadratic polynomial is f(x)= k ( x² - 2x/3 + ⅓)
HOPE THIS ANSWER WILL HELP YOU….
Answer:
f(x)=x
2
−2x+3 have zeroes α,β
⇒α+β=2
⇒α⋅β=3
Now polynomial having α+2,β+2 as roots is
p(x)=x
2
−(α+2+β+2)x+(α+2)(β+2)
=x
2
−(α+β+4)x+αβ+2(α+β)+4
=x
2
−(2+4)x+3+2(2)+4
⇒x
2
−6x+11