Are extraneous solution always negative numbers
2 answers:
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Answer:
x = - 1 ± 2i
Step-by-step explanation:
we can use the discriminant b² - 4ac to determine the nature of the roots
• If b² - 4ac > , roots are real and distinct
• If b² - 4ac = 0, roots are real and equal
• If b² - ac < 0, roots are not real
for x² + 2x + 5 = 0
with a = 1, b = 2 and c = 5, then
b² - 4ac = 2² - (4 × 1 × 5 ) = 4 - 20 = - 16
since b² - 4ac < 0 there are 2 complex roots
using the quadratic formula to calculate the roots
x = ( - 2 ± ) / 2
= (- 2 ± 4i ) / 2 = - 1 ± 2i