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1. Anyone have any ideas how I could approach this?

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.
2. (Original post by that_guy874)
Anyone have any ideas how I could approach this?

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
3. 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.

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