R richz Active Member Joined Sep 11, 2004 Messages 1,348 Jul 24, 2005 #1 for wat values of m does the equation x^3 - 12x^2 + 45x - m = 0 have three distinct solutions? thnx
O Ogden_Nash Member Joined Apr 27, 2004 Messages 35 Gender Male HSC 2005 Jul 24, 2005 #2 Let f(x) = x^3 - 12x^2 + 45x then find the stationary points of y = f(x). Then work out how far you can bring down the graph so that the stationary points are on opposite sides of the x-axis (ie 3 distinct roots). This will give you the values of m.
Let f(x) = x^3 - 12x^2 + 45x then find the stationary points of y = f(x). Then work out how far you can bring down the graph so that the stationary points are on opposite sides of the x-axis (ie 3 distinct roots). This will give you the values of m.
Slidey But pieces of what? Joined Jun 12, 2004 Messages 6,600 Gender Male HSC 2005 Jul 24, 2005 #3 That is: x^3 - 12x^2 + 45x = m m'=3x^2-24x+45=0 x^2-8x+15=0 x = 3, 5 m(3)=54 m(5)=50 Thus 51<=m<=53. If it were 50<=m<=54, you would be including when m gives only 2 roots (multiplicity). Last edited: Jul 24, 2005
That is: x^3 - 12x^2 + 45x = m m'=3x^2-24x+45=0 x^2-8x+15=0 x = 3, 5 m(3)=54 m(5)=50 Thus 51<=m<=53. If it were 50<=m<=54, you would be including when m gives only 2 roots (multiplicity).