For what real values of does the quadratic 12x^2+k*x+27=0 have nonreal roots

Step-by-step explanation:

Discriminant : k² - 4(12)(27) < 0

So k² - 1296 < 0, -36 < k < 36.

If a quadratic has nonreal roots, then the discriminant .  b^2 - 4ac

Plug in the values for b, a, and c into this equation to get .  k^2-4 *12*27 < 0

Rearrange to get .  k^2 < *12*27

Now factorize 4*12*27 and solve to get - 36 < k<26

Step-by-step explanation:

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answer:

the position of equilibrium will shift to the left.

step-by-step explanation:

the general equation for the dissociation of a weak base is

b + h₂o   ⇌   bh⁺ + oh⁻

according to le châtelier's principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

if you apply a stress by adding more bh⁺ or oh⁻,   the system will respond by moving in a direction that gets rid of the added ions.

the position of equilibrium will shift to the left, forming more of the undissociated base.

5+2x=a

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