Question

Find the values of k such that f(x)=x^2+kx+k+3 has no real roots.

Answer 1

hint :
D < 0

D = discriminant = b^2 - 4ac

Answer 2

did you get it, @rawritsmeli ?

Answer 3

Could the answer be (0, squareroot 12) -thats interval notation btw...

Answer 4

@RadEn

Answer 5

f(x)=x^2+kx+k+3
here, a = 1, b = k, and c = k+3
now, the value of discriminant is
D = b^2 - 4ac = k^2 - 4(1)(k+3) = k^2 - 4k - 12

Answer 6

D < 0, it means
k^2 - 4k - 12 < 0
(k -6)(k + 2) < 0

the value of k which makes the zeroes is
k = 6 or k = -2

graph the lines of interval for k and give the sign negative or positive values for k
++++(-2)--------(6)++++++

chose the negative area, because the problem is < 0, so the solution for k be -2<k<6

Answer 7

does that make sense ?? @rawritsmeli

Answer 8

Woah!! yes!
ok so i totally didn't think that k+3=c

Answer 9

ok, just remember that the general equation of function quadratic is
y = ax^2 + bx + c

Answer 10

or
f(x) = ax^2 + bx + c