Question

if alpha beta are the zero of the quadratic polynomial f(x)=x^2-x-2 , find a polynomial whose zeroes are 2alpha +1, 2 beta+1​

Answer 1

For polynomial f(x)=ax2+bx+cf(x)=ax2+bx+c

 Sum of roots =−ba=−ba

 Product of roots =ca=ca

f(x)=x2−x−2f(x)=x2−x−2

 Roots =α,β=α,β

α+β=1α+β=1

α×β=−2α×β=−2

g(x)=ax2+bx+cg(x)=ax2+bx+c

 Roots =2α+1,2β+1=2α+1,2β+1

 Sum of roots =(2α+1)+(2β+1)=2(α+β)+2=2(1)+2=4=(2α+1)+(2β+1)=2(α+β)+2=2(1)+2=4

 Product of roots =(2α+1)(2β+1)=4αβ+2(α+β)+1=(2α+1)(2β+1)=4αβ+2(α+β)+1 
=4(−2)+2(1)+1=−5=4(−2)+2(1)+1=−5

g(x)=x2−4x−5g(x)=x2−4x−5

Alternate method:

g(x)=x2−x−2=0g(x)=x2−x−2=0

(x−2)(x+1)=0(x−2)(x+1)=0

x=2,−1x=2,−1

g(x)g(x) has roots 2(2)+1=52(2)+1=5 and 2(−1)+1=−12(−1)+1=−1

g(x)=(x−5)(x+1)g(x)=(x−5)(x+1)

g(x)=x2−4x−5g(x)=x2−4x−5

Answer 2

Answer:

X^2 -4X -5 = 0

Step-by-step explanation:

2alpha +1, 2 beta+1​ are root so

(X - 1 -2alhpa)(X-1 -2beta) = 0

(X-1)^2 -2(X-1)(alhpa + beta) + 4alphabeta = 0

X^2 +1 - 2X -2X(alpha+beta) +2(alpha+beta) + 4alphabeta = 0

Now getting alpha+ beta and alphabeta from given equation

X^2 - X - 2 = 0

Alpha + beta = 1

Alphabeta = -2

Putting these values in equation

X^2 +1 - 2X -2X(1) +2(1) + 4(-2) = 0

X^2 -4X -5 = 0