Question

If one zero of the polynomial ax^2+bx+c is double the other,prove that 2b^2=9ac

Answer 1

let two zeroes are A and B

given A = 2 B

sum of zeros A + B = - b / c

2B + B = - b / c

3B = - b / c

B = - b / 3c ------------------(1)

product of zeros AB = c / a

2B×B = c / a

2B^2 = c / a

2( - b /3 c ) ^2 = c /a

2b^2 / 9c^2 = c /a

2b^2 = 9c^3 / a

"hope it helps "