Question

Show that the roots of the equation (a + b)x² + (b - c) x - (a - b + c) = 0 are always real if those
of ax² + 2bx + c = 0 are imaginary.

Answer 1

Step-by-step explanation:

Roots of ax² + 2bx + c are imaginary means its discriminant is less than 0. Means,

=> (2b)² - 4(a)(c) < 0

=> 4b² - 4ac < 0

=> 0 < 4ac - 4b²

Means, 4ac - 4b² is greater than 0.

Now, discriminant of the 2nd equation:

=> (b - c)² - 4(a + b)(a - b + c)

=> b² + c² - 2bc + 4(a² - ab + ac + ab - b² + bc)

=> b² + c² - 2bc + 4(a² + ac - b² + bc)

=> b² + c² - 2bc + 4a² + 4ac - 4b² + 4bc

=> b² + c² + 2bc + 4a² + 4ac - 4b²

=> (b + c)² + 4a² + 4ac - 4b²

(b + c)² & 4a² are squares, it means they will never be -ve. Also, as we saw, 4ac - 4b² is always greater than 0.

=> all numbers are +ve, in special cases they can be 0, but not -ve.

Since discriminant is +ve & not -ve, roots will be real.

Hence it can be said that the roots of the equation (a + b)x² + (b - c) x - (a - b + c) = 0 are always real if those of ax² + 2bx + c = 0 are imaginary.