How do you solve # (x^2+5)/(2x)=5#?

Question

1188 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-30T19:37:19

#"x_1=5-2sqrt5#
#x_2=5+2sqrt5#

#(x^2+5)/(2x)=5#

#x^2+5=2x*5#

#x^2+5=10x#

#x^2-10x+5=0#

#"This is a quadratic function .We should find the roots of the function."#

#"if "ax^2+bx+c=0#

#Delta=sqrt (b^2-4ac)#

#"Where "a=1" , "b=-10" , "c=5#

#Delta=sqrt((-10)^2-4*1*5)=sqrt(100-20)=+-sqrt(80)#

#x_1=(-b-Delta)/(2a)=(10-sqrt 80)/(2*1)=(10-4sqrt5)/2=5-2sqrt5#

#x_1=(-b+Delta)/(2a)=(10+sqrt 80)/(2*1)=(10+4sqrt5)/2=5+2sqrt5#