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

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neela | High School Teacher | (Level 3) Valedictorian

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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.

 

 

 

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