Question

Show: x2 + 1 has no real roots

Answer 1

For example, z2+1 has no real zeros (because its two zeros are not real numbers). x2-2 has no rational zeros (its two zeros are irrational numbers). The sine function has no algebraic zeros except 0, but has infinitely many transcendental zeros: -3pi, -2pi, -pi, pi, 2pi, 3pi etc.

Answer 2

Answer:

Step-by-step explanation:

Ok buddy here it comes to the question de +1 has no real roots cauz its ('DISCRIMINANT' is < 0) so its real part is zero and only non real or imaginary part exists