SOLUTION: Given y = −3x^2 + kx − 12 for k > 0, find the value(s) of k such that y < 0 for all x.

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Question 1177934: Given y = −3x^2 + kx − 12 for k > 0, find the value(s) of k such that y < 0 for all x.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

y+=+-3x%5E2+%2B+kx+-12+
for k+%3E+0, find the value(s) of k such that y+%3C+0 for all x.
y=-3x%5E2+%2B+kx+-12+
in the standard form
y=ax%5E2%2Bbx%2Bc+
a=-3, b=k, c=-12
discriminant:
b%5E2-4ac=k%5E2-4%28-3%29%28-12%29=k%5E2-4%2836%29=k%5E2-144
if k%5E2-144=0=>k=sqrt%28144%29=>k=12 or k=-12
the value(s) of k such that y+%3C+0 for all x:

and solution is: -12%3Ck%3C12+
or set { -11,-10,-9,.........,9,10,11 }


Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.

You are given that the quadratic function is always negative, for all values of x.


It means that it has no real roots.


In turn, it means that the discriminant of this quadratic function is negative.


The discriminant of a quadratic function  y = ax^2 + bx + c is


    d = b^2 - 4ac.


In your case a= -3,  b= k,  c= -12,  so the discriminant is


    d = k^2 - 4*(-3)*(-12) = k^2 - 144.


You should determine "k" from this condition


    d < 0,   or  k^2 - 144 < 0.


The solution is  k^2 < 144,   or   -12 < k < 12.


ANSWER.  The values of "k"  are   -12 < k < 12.


Solved.