Question

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

Answer 1

Answer:

b < - 2 or 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 or b > 2