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28 October, 13:04

4.

The roots of the equation x2 + 16 = 4bx are real if :

+2
Answers (1)
  1. 28 October, 15:48
    0
    b 2

    Step-by-step explanation:

    Given

    x² + 16 = 4bx (subtract 4bx from both sides)

    x² - 4bx + 16 = 0 ← in standard form

    with a = 1, b = - 4b, c = 16

    For the roots to be real the discriminant b² - 4ac > 0, that is

    ( - 4b) ² - (4 * 1 * 16) > 0

    16b² - 64 > 0

    16 (b² - 4) > 0 ← factor using difference of squares

    16 (b - 2) (b + 2) > 0, thus

    b 2
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