What is the value of #k# ?

Question

Find the value of #color(red)k# for which the inequality #color(red)(x^2-2(4k-1)x+15k^2-2k-7>0# is valid for ANY value of #color(red)x#

1542 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-16T18:44:07

The answer is #k in (2,4)#

For a quadratic equation #f(x)>0#, the graph must touch the x-axis

The discriminant must be #=0#

#f(x)=x^2-2(4k-1)x+15k^2-2k-7#

#Delta=(2(4k-1))^2-4(1)(15k^2-2k-7)=0#

#=>#, #4(16k^2-8k+1)-60k^2+8k+28=0#

#=>#, #64k^2-32k+4-60k^2+8k+28=0#

#=>#, #4k^2-24k+32=0#

#=>#, #4(k^2-6k+8)=0#

#=>#, #(k-2)(k-4)=0#

#=>#, #{(k=2),(k=4):}#

When #k=2#

#=>#, #x^2-14x+49=0#

graph{x^2-14x+49 [-14.48, 21.55, -4.62, 13.4]}

When #k=4#

#=>#, #x^2-30x+225=0#

graph{x^2-30x+225 [-14.48, 21.55, -4.62, 13.4]}

When #k=3#

#=>#, #x^2-22x+122=0#

graph{x^2-22x+122 [-14.48, 21.55, -4.62, 13.4]}