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

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Answer 1 · 2017-04-30T19:37:19

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

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

$x^2+5=2x*5$

$x^2+5=10x$

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

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

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

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

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

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

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

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

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