Solve k= k+15 divided by 3(k-1) use quadratic formula?
Answer:
k = (k+15)/3*(k-1)
==> multiply both sides by 3(k-1) to get rid of fractions
3k(k-1) = k+15
==> expand the LHS
3k^2-3k = k+15
==> move everything to the LHS
3k^2-4k-15=0
actually this does factor into (3k+5)(k-3), but if you want to use the formula, just use:
a=3, b=-4 and c=-15
x = -(-4) +/- sqrt[(-4)^2-2*3*(-15)] / 2*3
x = 4 +/- sqrt(196) / 2*3
x = 4 +/- 14 / 6
==> factor out a 2 from top and bottom
x = 2 (2 +/- 7) / 2*3
==> cancel the twos to get
x = 2 +/- 7 / 3
which gives 2+7 / 3 and 2-7 / 3 which gives 3 and -5/3
multiply by 3(k-1)
3k^2 - 3k = k + 15
3k^2 - 4k + 15 = 0
quadratic equation
x = (-b +/- sqrt(b^2 - 4*a*c)) / (2a)
answers are
x1 = 0.66667 + 2.1344 i
x2 = 0.66667 - 2.1344 i
you're right math, missed a sign change on 15
multiply by 3(k-1)
3k(k-1)=k+15
3k^2-3k-k-15=0
3k^2-4k-15=0
k=3 or k=-1.67
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