J jshaziza Junior Member Joined Jan 26, 2007 Messages 102 Jul 31, 2007 #1 N=(k^2-3k)/2 solve for k. Could someone show me at least one or two of the first steps in solving this. Thx.
N=(k^2-3k)/2 solve for k. Could someone show me at least one or two of the first steps in solving this. Thx.
M Mrspi Senior Member Joined Dec 17, 2005 Messages 2,116 Jul 31, 2007 #2 jshaziza said: N=(k^2-3k)/2 solve for k. Could someone show me at least one or two of the first steps in solving this. Thx. Click to expand... Multiply both sides by 2: 2N = k<SUP>2</SUP> - 3k Subtract 2N from both sides of the equation: 0 = k<SUP>2</SUP> - 3k - 2N Now....use the quadratic formula to solve the equation for k in terms of N..... a = 1 b = -3 c = -2N
jshaziza said: N=(k^2-3k)/2 solve for k. Could someone show me at least one or two of the first steps in solving this. Thx. Click to expand... Multiply both sides by 2: 2N = k<SUP>2</SUP> - 3k Subtract 2N from both sides of the equation: 0 = k<SUP>2</SUP> - 3k - 2N Now....use the quadratic formula to solve the equation for k in terms of N..... a = 1 b = -3 c = -2N
