The equation kx^2 + 4x + (5-k) = 0 , where k is a constant, has 2 different real solutions for x. a) Show that k satisfies the equation: k^2 - 5k + 4 > 0
The equation kx^2 + 4x + (5-k) = 0 , where k is a constant, has 2 different real solutions for x. a) Show that k satisfies the equation: k^2 - 5k + 4 > 0
b) Hence find the set of possible values for k.
Use the quadratic formula b^2 - 4ac A is k B is 4 C is 5-K
If that quadratic is strictly more than 0 then it means it never crosses the x- axis, which means it has no roots, you should be able to figure out the rest.