Question

If df-(e/2)^2 > 0, then the equation (ax^2+bx-a) (dx^2+ex+f) = 0 ( where a,b,c,d,e and f are real numbers) will have​

Answer 1

Given :  df-(e/2)² > 0

(ax²+bx-a) (dx²+ex+f) = 0

To Find :  (ax²+bx-a) (dx²+ex+f) = 0 will have roots

Solution:

(ax²+bx-a) (dx²+ex+f) = 0

ax²+bx-a = 0

D = b²  - 4(a)(-a) = b² + 4a²  > 0    

=> Roots are real and unequal

dx²+ex+f = 0

=> D = e²  - 4df

df-(e/2)² > 0   => 4df - e²  > 0

=> e² - 4df  < 0

D < 0

Hence roots are imaginary

(ax²+bx-a) (dx²+ex+f) = 0  will have

2 real and 2 imaginary roots

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