Algebra II - Quadratic Equations?

Question

I need to find all values of "k" such that x² + kx + 1 = 0 has two complex roots. Please help!

Answer 1

to have a complex root, b^2 - 4ac <0 since to be complex, you must have a negative number under a square root symbol
a=1, b = k , c = 1
b^2 - 4ac <0
k^2-4*1*1 <0
k^2-4<0
k^2<4
k<2, k>-2

-2

Answer 2

From the quadratic equation, you will have complex roots if
√(k^2 - 4) < 0
k^2 - 4 < 0
k^2 < 4
|k| < 2
-2 < k < 2

Answer 3

1st flow the 13 to different edge x^2 +14x +____= -13 +____ you want to leave an area on both edge for the variety that completes the sq.. (you may want to do an identical area of each edge of the = to maintain it balanced. discover C to fill interior the lacking #'s : c = (14/2)^2 = 7^2 = 40 9. Now plug it into the areas: x^2 +14x +40 9 = -13 + 40 9 now element the left edge, and simplify the right edge: (x+7)(x+7)=36 (x+7)^2 = 36 next take the sq. root of each edge: x+7 = 6,-6 subtract 7 from both a threat solutions: x = -7+6, -7-6 x=-a million,-13 2. This time you should divide each little thing by potential of 5 first: 5x^2 - 12x = -2 x^2 - 12/5x = -2/5 x^2 - 12/5x +___= -2/5 + ___ ; c = ((-12/5)/2)^2 = (-6/5)^2 = 36/25 x^2 - 12/5x +36/25= -2/5 + 36/25 element the left edge: (x-6/5)^2 = -2/5 +36/25 Get uncomplicated denominator on the right edge: -2/5 +36/25 =-10/25 + 36/25 = 26/25 (x-6/5)^2 =26/25 sq. root: x-6/5 = +/-sqrt(26)/5 upload -6/5: 6/5 +/- sqrt(26)/5

Answer 4

ax² + kx + c = 0
Examine possible values of m=k^2-4ac.
If m>0, two real roots, m=0, one real root, m<0, two complex roots.
So, k^2-4<0, k^2<4, -2

Answer 5

k^2-4<0
k^2<4
Therefore, when k is less then 2 and greater then -2 there will be two complex number solutions.

Answer 6

b^2 - 4ac < 0
so . .
k^2 - 4 < 0
so
k > -2 and k < 2