What are the value of #k#, for which #x^2+5kx+16=0# has no real roots?

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Answer 1 · 2016-11-02T07:08:06

The values of k for which the equation #x^2+5kx+16=0# has no real roots are #-8/5 < k < 8/5#

An equation #ax^2+bx+c=0# has no real roots if the determinant #b^2-4ac<0#

#x^2+5kx+16=0# has no real roots if the determinant #(5k)^2-4xx1xx16=25k^2-64<0# or

#(5k-8)(5k+8)<0#

As #(5k-8)(5k+8)# is negative,

either #5k+8<0# and #5k-8>0#

But this is just not possible, as #k# cannot be #k<-8/5# as well as #k>8/5#

Other alternative is #5k+8>0# and #5k-8<0# i.e.

#k> -8/5# as well as #k<8/5#. This is possible if value of #k# is between #-8/5# and #8/5#.

Hence #-8/5 < k < 8/5#