How do you solve #2x^2-8x+5>=0#?

Question

3238 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-31T06:25:38

Solution is #x>=2+sqrt(3/2)# or #x<=2-sqrt(3/2)#

#2x^2-8x+5>=0# is equivalent to

#2(x^2-4x+5/2)>=0# or

#2(x^2-4x+4-4+5/2)>=0# or

#2((x-2)^2-3/2)>=0# or

#(x-2)^2-(sqrt(3/2))^2>=0# or

#(x-2+sqrt(3/2))(x-2-sqrt(3/2))>=0# or

i.e. either #(x-2+sqrt(3/2))>=0# and #(x-2-sqrt(3/2))>=0# i.e.

#x>=2-sqrt(3/2)# and #x>=2+sqrt(3/2)# i.e. #x>=2+sqrt(3/2)#

or #(x-2+sqrt(3/2))<=0# and #(x-2-sqrt(3/2))<=0# i.e.

#x<=2-sqrt(3/2)# and #x<=2+sqrt(3/2)# i.e. #x<=2-sqrt(3/2)#

Hence solution is #x>=2+sqrt(3/2)# or #x<=2-sqrt(3/2)#

graph{2x^2-8x+5 [-3, 7, -3.16, 1.84]}