Question

where k is a real constant.

a.   Find the discriminant of f(x) in terms of k.

b.   Show that the discriminant of f (x) can be expressed in the form (k + a)² + b, where a and b are  integers to be found.

c.   Show that, for all values of k, the equation f (x) = 0 has real roots.

Solution

a.
 

We are given;

It is evident that we are given a quadratic equation and we are required to find discriminant of this  equation.

For a quadratic equation , the expression for solution is;

Where  is called discriminant.

Therefore;

b.
 

We have found discriminant of given quadratic equation in (a) as;

We have the expression;

We use method of “completing square” to obtain the desired form.

Next we complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore we can deduce that;

Hence we can write;

To complete the square we can add and subtract the deduced value of ;

c.
 

We are given that f(x)= has real roots.

Since ;

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 we are given only equation has real roots but not that these roots are distinct or otherwise.

Therefore, for this equation;

We have found in (a) that discriminant of this equation is;

Therefore, we need to solve the inequality;

We have also demonstrated in (b) that  can be written as;

Hence;

It is evident that for all real values of k the above equation will be;

Hence, the given equation will have real roots for all real values of k.