How do you solve #x+1/x=7#?

2 Answers
May 25, 2018

#color(red)(x=3.85 or 0.15#

Explanation:

#x+1/x=7#

or, #x+1/x-7=0#

or, #(x^2+1-7x)/x=0#

or, #x^2-7x+1=0#

or, #x=(-(-7)+-sqrt((7^2)-4*1))/(2*1)=(7+-sqrt45)/2=(7+-3sqrt5)/2#

#therefore x=3.5+1.5sqrt5 or 3.5-1.5sqrt5#

or, #x=3.5+3.35 or 3.5-3.35#

or, #x=3.85 or 0.15#

May 25, 2018

#x=7/2+-3/2sqrt5#

Explanation:

#"multiply through by "x#

#x^2+1=7x#

#"arrange in "color(blue)"standard form ";ax^2+bx+c=0#

#"subtract "7x" from both sides"#

#x^2-7x+1=0#

#"check the value of the "color(blue)"discriminant"#

#•color(white)(x)Delta=b^2-4ac#

#"with "a=1,b=-7" and "c=1#

#b^2-4ac=(-7)^2-(4xx1xx1)=45#

#"thus the roots are real and irrational"#

#"solve using the "color(blue)"quadratic formula"#

#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(7+-sqrt45)/2=(7+-3sqrt5)/2#

#x=7/2+-3/2sqrt5larrcolor(red)"exact solutions"#