The equation `a x^2-b x+c=0` has real and positive roots. Prove that the roots of the equation `a d^2x^2+a(3b-2c)x+(2b-c)(b-c)+a c=0` re real and positive.

Explanation

Given `ax^(2) + bx + c = 0` has real and positive roots . Then
`b^(2) - 4ac ge` 0 (1)
Sum of roots is
`-b//a gt or b//a lt 0` (2)
i.e., a and c have oppsote sign.
Product of root is
`c//a gt 0` (3)
i.e., a and c have same sign.
Now, for equation `a^(2) x^(2) + a(3b - 2c) x + (2b - c) (b - c) + ac = 0` ,
we have
D `= a^(2) (3b - 2c)^(2)-4a^(2)[(2b - c)(b - c) +ac]`
`= a^(2) [9b^(2) - 12bc +4c^(2) - 4(2b^(2) - 3bc + c^(2) + ac)]`
`= a^(2) [9b^(2) - 12bc +4c^(2) - 8b^(2) - 12bc - 4c^(2) + 4ac]`
` = a^(2)( b^(2) - 4ac) ge 0` [ Using (1)]
Hence, the roots are real. Also, sum of roots,
`(-a(3b -2c))/(a^(2)) = -((3b)/(a) - (2c)/(a)) gt 0` [ Using (2) and (3)]
Product of roots,
`((2b -c)(b -c) + ac)/(a^(2)) = (2 (b)/(a) - (c)/(a))((b)/(a)-(c)/(a)) + (c)/(a) gt 0` [Using (2) and (3)]
Hence, the roots are positive.