Find the value of k if f(x) has three distinct real roots ?

If equation has three real roots all of them are distinct, then

#f(x)=(x-a)(x-b)(x-c)# where #a,b, c# are roots of #f(x)#

Developing:

#f(x)=(x^2-bx-ax+ab)(x-c)=x^3-cx^2-bx^2+bcx-ax^2+acx+abx-abc=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc#

We know that two polynomial expresions are equal if and only if his coefficients are equal. Then (Cardano formulae)

#a+b+c=0#
#ab+ac+bc=-3#
#-abc=k#

Now, if we take a look to #x^3-3x# we see that has 3 roots if we re-write in form #x(x^2-3)# ; #x=0, x=+-sqrt3#

Adding +k we are traslating graph up or down if k is positive or negative respectively and this fact we can eliminate 2 real roots adding 2 complex roots.(if a polinomial has a complex root, then his conjugate is also root). Hope this helps

Let's start by taking a look at #x^3-3x#: it has three roots, and local minimum and a maximum:

graph{x^3-3x [-2 2 -3 3]}

The minimum and maximum are found by posing

#f'(x)=3x^2-3 = 0 \iff x^2-1=0 \iff x=\pm 1#

With corresponding values #f(1)=1-3=-2# for the minimum and #f(-1) = -1+3=2# for the maximum.

Adding #k# to the function translates the graph upwards (#k>0#) or downwards (#k<0#). You can see that the function has three zeroes as long as the maximum is above the #x# axis, and the minimum is below.

If we translate the function up three units, for example, the minimum would be above the #x# axis, and we would lose solutions:

graph{x^3-3x+3 [-5 5 -1 6]}

Similarly, we can't translate the graph down more than two units: see this example with #k=-3#

graph{x^3-3x-3 [-5 5 -6 1]}

While any #k# between #-2# and #2# is ok:

graph{x^3-3x+1 [-3 3 -2 4]}