Find any number which when added to its reciprocal gives 5.

[TheUprightGuy]

Self explaining title, find any number which when added to its reciprocal gives 5.
This presumably would be x+1/x = 5
Putting this to quadratics gets x²-5x+1=0
But putting this into the quadratic formula gets a long string of decimals that don't seem to work.
Any help?

 

[MarkFL]

I agree that the statement of the problem gives rise to the equation:

\(\displaystyle \displaystyle x+\frac{1}{x}=5\)

Which indeed leads to the quadratic:

\(\displaystyle \displaystyle x^2-5x+1=0\)

Use of the quadratic formula gives us:

\(\displaystyle \displaystyle x=\frac{5\pm\sqrt{21}}{2}\)

So, let's substitute these solutions into the original equation and see if they work:

\(\displaystyle \displaystyle \frac{5\pm\sqrt{21}}{2}+\frac{1}{\dfrac{5\pm\sqrt{21}}{2}}=5\)

\(\displaystyle \displaystyle \frac{5\pm\sqrt{21}}{2}+\frac{2}{5\pm\sqrt{21}} \cdot\frac{5\mp\sqrt{21}}{5\mp\sqrt{21}}=5\)

\(\displaystyle \displaystyle \frac{5\pm\sqrt{21}+5\mp\sqrt{21}}{2}=5\)

\(\displaystyle \displaystyle \frac{10}{2}=5\)

\(\displaystyle \displaystyle 5=5\quad\checkmark\)

So, yes, both solutions satisfy the condition that each, when added to its reciprocal, gives 5.

 

[TheUprightGuy]

I agree that the statement of the problem gives rise to the equation:

\(\displaystyle \displaystyle x+\frac{1}{x}=5\)

Which indeed leads to the quadratic:

\(\displaystyle \displaystyle x^2-5x+1=0\)

Use of the quadratic formula gives us:

\(\displaystyle \displaystyle x=\frac{5\pm\sqrt{21}}{2}\)

So, let's substitute these solutions into the original equation and see if they work:

\(\displaystyle \displaystyle \frac{5\pm\sqrt{21}}{2}+\frac{1}{\dfrac{5\pm\sqrt{21}}{2}}=5\)

\(\displaystyle \displaystyle \frac{5\pm\sqrt{21}}{2}+\frac{2}{5\pm\sqrt{21}} \cdot\frac{5\mp\sqrt{21}}{5\mp\sqrt{21}}=5\)

\(\displaystyle \displaystyle \frac{5\pm\sqrt{21}+5\mp\sqrt{21}}{2}=5\)

\(\displaystyle \displaystyle \frac{10}{2}=5\)

\(\displaystyle \displaystyle 5=5\quad\checkmark\)

So, yes, both solutions satisfy the condition that each, when added to its reciprocal, gives 5.

Whilst I see what you're doing to try to get to it, try putting 5 in place of x in the initial formula.

\(\displaystyle \displaystyle 5+\frac{1}{5}=5\)

Unless I am really rusty with my mathes,
\(\displaystyle \displaystyle 5+\frac{1}{5}=5\frac{1}{5}\)
So therefore I don't think 5 is the solution

 

[HallsofIvy]

Yes, "5" is not a solution and MarkFL did not say it is!

MarkFL said that \(\displaystyle x= \frac{5+ \sqrt{21}}{2}\) and \(\displaystyle x= \frac{5- \sqrt{21}}{2}\) are solutions to this equation

because they both make \(\displaystyle x+ \frac{1}{x}\) equal to 5.

 

[TheUprightGuy]

Yes, "5" is not a solution and MarkFL did not say it is!

MarkFL said that \(\displaystyle x= \frac{5+ \sqrt{21}}{2}\) and \(\displaystyle x= \frac{5- \sqrt{21}}{2}\) are solutions to this equation

because they both make \(\displaystyle x+ \frac{1}{x}\) equal to 5.

Ohhhh. Understood. I guess I expected the answer to be a whole number. previous question from the lecturer have proven otherwise.
Thank you so much for the help, the feeling of the sudden dawning of understanding is why I have a love/hate relationship with mathes