Given f(x) = kx^2-4kx+16, what is the value of k so f(x) = 0 has: a) no real roots, b) two equal roots, c) two distinct roots?

1 Answer
Jun 18, 2018

#cc a) k in [0,4]#
#cc b) k in {0,4}#
#cc c) k in (-oo,0) uu (4,+oo)#

Explanation:

We have

#f(x)=kx^2-4kx+16#

The roots of the equation are given by the discriminant #Delta#:

#Delta = b^2-4ac#

#{(x_1 =(-b+sqrtDelta)/(2a)),(x_2=(-b-sqrtDelta)/(2a)) :}#

As such, the conditions of existence of #x_1# and #x_2# are determined by #Delta#.

#diamond Delta>0 " there are two distinct real solutions"#
#diamond Delta=0 " there is one solution"#
#diamond Delta<0 " there are no solutions in real numbers"#

#Delta = 16k^2-64k#

#cc a)" f(x) has no real roots"#

#:. Delta < 0 => 16k^2-64k<0#

For a quadratic function #phi# with leading coefficient positive and roots #r_1#, #r_2#, #r_1>r_2#, the equation

#phi(x)<0#

has solutions in the interval

#x in (r_2,r_1)#

In our case,

#phi(k) = 16k^2-64k=16k(k-4)#

#{(r_1 = 4), (r_2 = 0) :}#

#phi(k)< 0 => color(red)(k in (0,4)#

#cc b)" f(x) has two equal roots"#

#:. Delta = 0 => 16k^2-64k=0 => color(red)(k in {0,4}#

#cc c)" f(x) has two distinct roots"#

#:. Delta > 0 => 16k^2-64k>0#

Similarly, the equation

#phi(x)>0#

where #phi# is the same quadratic function has solutions

#x in (-oo, r_2) uu (r_1, +oo)#

#phi(k) > 0 => 16k^2-64k > 0 => color(red)(k in (-oo, 0) uu (4, + oo)#