Show that the solutions of the equation x^2+kx= 3-k are real for all real values of k
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use discriminat.
x^2 + kx +(k-3) = 0 has discriminant = k^2 - 4(k-3) = k^2 + 4k +12 = (k+2)^2 + 8 which cannot be - ve for any k. hence etc.
x^2 + kx +(k-3) = 0 has discriminant = k^2 - 4(k-3) = k^2 + 4k +12 = (k+2)^2 + 8 which cannot be - ve for any k. hence etc.