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direct variation, and in the equation y=kx the k ca NOT equal 0.
direct variation: y = kx y = kx k = y/x = 0.8/0.4 = 2
Using the discriminant for a quadratic equation the value of k works out as plus or minus 12.
y = kx ie y/x = k, k = 2, so y = 2x
If: kx+y = 4 and y = x^2 +8 Then: x^2 +8 = 4-kx or x^2 +8 -4+kx = 0 => x^2+4+kx = 0 The discriminant of the above quadratic equation must equal 0 So: k^2 -4*(4*1) = 0 => k^2-16 = 0 Therefore: k^2 = 16 and so the values of k are -4 and +4