How do you solve the equation: #x + 1/x+1 = 4#?

2 Answers
Jul 23, 2015

#x=\frac{3\pm sqrt{5}}{2}\approx 2.618, 0.382#

Explanation:

Two methods:

1) Multiply everything by #x# first to get #x^2+1+x=4x#, or #x^2-3x+1=0#. The quadratic formula now gives #x=\frac{3\pm sqrt{9-4}}{2}=\frac{3\pm sqrt{5}}{2}\approx 2.618, 0.382#.

2) Add the fractions on the left first by getting a common denominator of #x#:

#x^2/x+1/x+x/x=4\Rightarrow (x^2+x+1)/x=4/1#

Now "cross-multiply" to get #x^2+x+1=4x#, or #x^2-3x+1=0#. This gives the same answer now as Method 1: #x=\frac{3\pm sqrt{9-4}}{2}=\frac{3\pm sqrt{5}}{2}\approx 2.618, 0.382#.

As with all algebra problems like this, you should check the answers in the original equation. If you try this with, for example, #x=(3+sqrt{5})/2# in exact form, you'll get:

#x+1/x+1=(3+sqrt{5})/2+2/(3+sqrt{5})+1#

#=((3+sqrt{5})(3+sqrt{5})+2*2+2(3+sqrt{5}))/(2(3+sqrt{5}))#

#=(9+6sqrt{5}+5+4+6+2sqrt{5})/(6+2sqrt{5})=(24+8sqrt{5})/(6+2sqrt{5})#

#=(4(6+2sqrt{5}))/(6+2sqrt{5})=(4cancel((6+2sqrt{5})))/(cancel(6+2sqrt{5}))=4#

Jul 23, 2015

Multiply all terms by #x#; rearrange and solve as a standard quadratic to get:
#color(white)("XXXX")##x = (3+-sqrt(5))/2#

Explanation:

#x + 1/x + 1 = 4#

#rArr##color(white)("XXXX")##x^2+1+1x = 4x#

#rArr##color(white)("XXXX")##x^2-3x+1 = 0#

using the quadratic formula:
#rArr##color(white)("XXXX")##x = (3+-sqrt((-3)^2-4(1)(1)))/(2(1)#

#color(white)("XXXX")##color(white)("XXXX")## x = (3+-sqrt(5))/2#