Question

if alpha and beta are zeroes of polynomial 6x²-7x-3, then form a quadratic polynomial where zeroes are 2alpha and 2beta.

Answer 1

6x^2-9x+2x-3
2x(3x+1) -3(3x+1)
(2x-3)(3x+1)
x = 3/2 , -1/3

Answer 2

6x^2-7x-3=0
Sum=α+ß
=-b/a
=-(-7)/6
=7/6
Product=αß
=c/a
=-3/6
=-1/2

Sum of new polynomial=1/α+1/ß
=(α+ß) {By substituting}
αß
=7/6+(-1/2)
7/6*(-1/2)
=7/6-3/6
7/6*(-1/2)
=4/6
-7/12
=-8/7
Product of new zeroes=1/α*1/ß
=1/(α*ß) {By substituting}
=1
-1/2
=-2

Required polynomial=k(x^2-[sum of new zeroes]x+[product of new zeroes])
=k(x^2-[-8/7]x+[-2]) {By substituting}
=k(x^2+8x/7-2)
=k(7x^2+8x-14)
7
=7x^2+8x-14(where k=7)

this is your answer....