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Let `f(x) = x^2+2x+9; x in RR` Determine the set of values of a real constant `lambda`...
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`f(x) = x^2+2x+9`
`x^2+2x+9 = lambda`
`x^2+2x+9-lambda = 0`
If the above quadratic function does not have real roots then the discriminant(Delta) should be less than 0.
`Delta = 4-4xx1xx(9-lambda)<0`
So for `f(x) = lambda` where there is no solutions `lambda<8`
Posted by jeew-m on September 18, 2013 at 2:18 AM (Answer #1)
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