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

Learn More:

Which expression is a possible leading term for the polynomial ...

brainly.in/question/13233517

18. Find the zeroes of the quadratic polynomial 4x2 – 6 – 8x and ...

brainly.in/question/3673291

Answer 2

$\huge\mathcal\colorbox{black}{{\color{white}{ANSWER :)}}}$

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

Learn More:

Which expression is a possible leading term for the polynomial ...

brainly.in/question/13233517

18. Find the zeroes of the quadratic polynomial 4x2 – 6 – 8x and ...

brainly.in/question/3673291