# For what values of k is x^2+kx+9> and equals to 0?

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Solve for k for which x^2+kx+9> and equals to 0?

Solution:

x^2+kx+9 = y(x)>=0 implies

y(x) = [(x+k/2)^2]-k^2/4+9 should be >=0

=> [(x+k/2)^2]+(-k^2/4+9) >=0

But (x+k/2)^2 is always >=9 as it is the square of a real value. Therefore, the other part -k^2/4+9 needs to be >= 0 or -k^2+36 needs to be 0 or psitive for y(x) to be positive.

=> k^2 needs to be less than or equal to 36.

Therefore, the value of k should lie between -6 and 6 or **-6<=k<=6** in order that the expression, x^2+kx+9 is positive always or for all values of x.