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

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.

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Eloise 3 weeks 2021-11-07T10:27:41+00:00 1 Answer 0 views 0

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    2021-11-07T10:29:23+00:00

    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.

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