# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2009 | January | Q#7

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Question

The equation kx+ 4x + (5 – k) = 0, where k is a constant, has 2 different real solutions for x.

a.   Show that k satisfies k2 – 5k + 4 > 0.

b.   Hence find the set of possible values of k.

Solution

a.

We are given;

We are given that given equation has two different real solutions of x (roots).

For a quadratic equation , the expression for solution is;

Where  is called discriminant.

If , the equation will have two distinct roots.

If , the equation will have two identical/repeated roots.

If , the equation will have no roots.

Since given is a quadratic equation with two different real solutions of x (roots), its discriminant must be;

b.

We are required to solve the inequality;

We solve the following equation to find critical values of ;

Now we have two options;

Hence the critical points on the curve for the given condition are 1 & 4.

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If  (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
If
(‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

We recognize that  it is an upwards opening parabola.

Therefore conditions for  are;