S Stine New member Joined Nov 29, 2006 Messages 49 Dec 10, 2006 #1 Find two values for k, for which the quadratic equation 4x^2 + kx + 81 = 0 has two equal real solutions Oh I am lost! :?
Find two values for k, for which the quadratic equation 4x^2 + kx + 81 = 0 has two equal real solutions Oh I am lost! :?
M Mrspi Senior Member Joined Dec 17, 2005 Messages 2,127 Dec 10, 2006 #2 Re: Find two values for k Stine said: Find two values for k, for which the quadratic equation 4x^2+kx+81=0 has two equal real solutions Oh I am lost! :? Click to expand... You haven't covered the discriminant in the quadratic formula yet? Most strange you'd be assigned this kind of problem........ The part of the quadratic formula which is under the radical sign, b<SUP>2</SUP> - 4ac, tells you what kind of roots a quadratic equation will have. if b<SUP>2</SUP> - 4ac = 0 the quadratic will have two equal real solutions. In your equation, a = 4, b = k, and c = 81. Substitute those values into b<SUP>2</SUP> - 4ac = 0 and solve for k.
Re: Find two values for k Stine said: Find two values for k, for which the quadratic equation 4x^2+kx+81=0 has two equal real solutions Oh I am lost! :? Click to expand... You haven't covered the discriminant in the quadratic formula yet? Most strange you'd be assigned this kind of problem........ The part of the quadratic formula which is under the radical sign, b<SUP>2</SUP> - 4ac, tells you what kind of roots a quadratic equation will have. if b<SUP>2</SUP> - 4ac = 0 the quadratic will have two equal real solutions. In your equation, a = 4, b = k, and c = 81. Substitute those values into b<SUP>2</SUP> - 4ac = 0 and solve for k.
