Answer:
Step-by-step explanation:
Hello,
I understand that the equation is
[tex]x+\dfrac{1}{x}=6[/tex]
If this is not the case please let me know quickly so that I can still change my answer
so first of call we assume that x is different from 0 as dividing by 0 is not allowed
and then we can multiply by both parts of the equation to get
[tex]x^2+1=6x\\<=>\\x^2-6x+1=0[/tex]
[tex]\Delta=b^2-4ac = 6^2-4=36-4=32=16*2\\\\x_1=\dfrac{6-\sqrt{\Delta}}{2}=\dfrac{6-\sqrt{16*2}}{2}=\dfrac{6-4\sqrt{2}}{2}=3-2\sqrt{2}\\\\x_2=\dfrac{6+\sqrt{\Delta}}{2}=3+2\sqrt{2}\\\\[/tex]
there are two solutions
Hope this helps
