Question

The zeroes of the polynomial f(x) = x^2+x+3/4 are​

Answer 1

Step-by-step explanation:

f(x) = x² + x + 3/4

in general, such a quadratic function is defined as

f(x) = a×x² + b×x + c

the solution for finding the values of x where a quadratic function value is 0 (there are as many solutions as the highest exponent of x, so 2 here in our case)

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 1

c = 3/4

x = (-1 ± sqrt(1² - 4×1×3/4))/(2×1) =

= (-1 ± sqrt(1 - 3))/2 = (-1 ± sqrt(-2))/2 =

= (-1 ± sqrt(2)i)/2

x1 = (-1 + sqrt(2)i) / 2

x2 = (-1 - sqrt(2)i) / 2

remember, i = sqrt(-1)

f(x) has no 0 results for x = real numbers.

for the solution we need to use imaginary numbers.

Answer 2

Answer:

• No real zeros

Step-by-step explanation:

I'll solve it by completing the square:

f(x) = x² + x + 3/4 =

        x² + 2*(1/2)x + (1/2)² - (1/2)² + 3/4 =

        (x + 1/2)² - 1/4 + 3/4 =

        (x + 1/2)² + 1/2

We got the expression, which is greater than zero for any value of x. It means the graph has no intersection with the x-axis, and there are no real zero's.