Find all real numbers #z# for which the equation #(z-5)x^2-zx+5=0# only has one real solution. Help, Please?

Question

3731 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-25T03:08:43

#=>z=10+-4sqrt(5)#

In order for a quadratic equation in the form #ax^2+bx+c=0# to have 1 solution, #b^2-4ac=0#

In this case:

plug in the values to get:

#(-z)^2-4(z-5)(5)=0#

#=>(-z)^2-4(5z-5)=0#

#=>z^2-20z+20=0#

#=>z=(-(-20)+-sqrt((-20)^2-4(1)(20)))/(2*1)#

#=>z=(20+-sqrt(400-80))/2#

#=>z=(20+-sqrt(320))/2#

#=>z=(20+-8sqrt(10))/2#

#=>z=10+-4sqrt(5)#