Consider this system of equations: y = x2 y = x + k For which value of k does the system have no real number solutions? For which value of k does the system have one real number solution? For which value of k does the system have two real number solutions?

We solve the system of equations:

y = x ^ 2

y = x + k

Resolving we have:

x ^ 2 = x + k

x ^ 2 - x - k = 0

We apply the resolver for the second degree polynomial:

x = (-b + / - root (b ^ 2 - 4 * a * c)) / 2 * a

We substitute the values:

x = ( - ( - 1) + / - root (( - 1) ^ 2 - 4 * (1) * ( - k))) / 2 * (1)

x = (1 + / - root (1 + 4 * (1) * (k))) / 2

no real number solutions:

1 + 4 * (1) * (k) <0

k <-1/4

one real number solution:

1 + 4 * (1) * (k) = 0

k = - 1/4

two real number solutions:

1 + 4 * (1) * (k) > 0

k> - 1/4

For which value of k does the system have no real number solutions? = - 2

for which value of k does the system have one real number solution? = - 0.25

for which value of k does the system have two real number solutions? = 2