Question 380121: If the roots of x^2+kx+3=0 are real, find the smallest integral value of k.
How would you solve this? Found 2 solutions by stanbon, richard1234:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! If the roots of x^2+kx+3=0 are real, find the smallest integral value of k.
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If the roots are real b^2-4ac > 0
k^2-4*1*3 > 0
k^2 > 12
k > 2sqrt(3) = 2.36...
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Smallest integral value of k = 3
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Cheers,
Stan H.
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You can put this solution on YOUR website! We want the discriminant, b^2 - 4ac, to be positive. Therefore we have
k^2 - 12 > 0
k^2 > 12
|k| >= 4
However, k can be positive or negative -- if k were negative, k can be negative infinity and the roots of the quadratic would be real. The smallest absolute value for k would be 4, though.
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