Question

N x² - 10x + 41 = 0
a) 1
b) 2
c) 3
d) 4

Answer 1

The answer is that the equation has two complex numbers as solutions.

The given quadratic equation is
`\(x^2 - 10x + 41 = 0\)`.

To determine the number of complex solutions, we can use the discriminant
`(\(\Delta\))` of the quadratic formula
`\(ax^2 + bx + c = 0\)`, where
`\(\Delta = b^2 - 4ac\).`

If
`(\(\Delta\))` is positive, the equation has two distinct real solutions.

If
`(\(\Delta\))` is zero, the equation has one real solution (a repeated or double root).

If
`(\(\Delta\))` is negative, the equation has two complex (conjugate) solutions.

In this case, the coefficients are a = 1 b = -10, and c = 41

Calculate
`(\(\Delta\))` :

`\[ \Delta = (-10)^2 - 4(1)(41) = 100 - 164 = -64 \`

Since
`\(\Delta\)` is negative
`(\(\Delta < 0\))`, the quadratic equation
`\(x^2 - 10x + 41 = 0\)`has two complex solutions.

Therefore, the answer is that the equation has two complex numbers as solutions.

How many complex numbers this equation has. x² - 10x + 41 = 0

a) 1

b) 2

c) 3

d) 4