The equation x^2 + 4 = 0 has no real root. What are its complex roots?
A2 and -2
B4i and -4i
C2i and -2i
D4 and -4
Answer & Solution
Correct answer: C. 2i and -2i
x squared = -4 gives x = plus or minus 2i, since (2i) squared equals -4.
Keep practising Complex Numbers and Quadratic Equations
Browse all Complex Numbers and Quadratic Equations questions for MHT-CET, or explore more Mathematics topics.
Related questions
What is the real part of (2 + i) squared?Evaluate (2 + 3i)(1 - i) and write the result in the form a + bi.Adding two complex numbers on the plane gives the fourth vertex of which shape?Geometrically, multiplying by a complex number does what to the plane?The complex numbers form a real vector space of what dimension?The polynomial x^2 - 2 has no root in which set?The roots of x^2 + 2x + 5 = 0 are which pair?Which number set contains a root of every non-constant polynomial equation?