Algebra problem

[smtchahal]

The number obtained by adding the squares of a number and its reciprocal is greater by 7/4 than the sum of the number and its reciprocal. Find the number.

 

[soroban]

Hello, smtchahal!

Both the set-up and the subsequent algebra are tricky.

The number obtained by adding the squares of a number and its reciprocal
is greater by 7/4 than the sum of the number and its reciprocal. . Find the number.

Let \(\displaystyle x\) = the number.
Then \(\displaystyle \frac{1}{x}\) = its reciprocal.

We are told: (sum of their squares) = (their sum) plus 7/4

So we have: .\(\displaystyle x^2 + \left(\dfrac{1}{x}\right)^2 \;=\;x + \dfrac{1}{x} + \dfrac{7}{4}\)

. . . . . . . . . . . . \(\displaystyle x^2 + \dfrac{1}{x^2} \;=\;x + \dfrac{1}{x} + \dfrac{7}{4}\)

Add 2 to both sides: .\(\displaystyle x^2 + 2 + \dfrac{1}{x^2} \;=\;x + \dfrac{1}{x} + \dfrac{15}{4}\)

. . . . . . . . . . . . . . . . .\(\displaystyle \left(x + \dfrac{1}{x}\right)^2 \;=\;x + \dfrac{1}{x} + \dfrac{15}{4}\)

. . . \(\displaystyle \left(x + \dfrac{1}{x}\right)^2 - \left(x + \dfrac{1}{x}\right) - \dfrac{15}{4} \;=\;0\)

. \(\displaystyle 4\left(x + \dfrac{1}{x}\right)^2 - 4\left(x + \dfrac{1}{x}\right) - 15 \;=\;0 \;\hdots\;\text{ a quadratic!}\)


Let \(\displaystyle u \,=\,x + \frac{1}{x}\!:\;\;4u^2 - 4u - 15 \:=\:0 \quad\Rightarrow\quad (2u + 3)(2u - 5) \:=\:0\)

. . Hence: .\(\displaystyle u \:=\:\text{-}\dfrac{3}{2},\;\dfrac{5}{2}\)


Back-substitute:

\(\displaystyle x + \dfrac{1}{x} \:=\:\text{-}\dfrac{3}{2} \quad\Rightarrow\quad 2x^2 + 3x + 2 \:=\:0 \quad\Rightarrow\quad x \:=\:\dfrac{-3\pm\sqrt{-7}}{4} \;\;?\)

\(\displaystyle x + \dfrac{1}{x} \:=\:\dfrac{5}{2} \quad\Rightarrow\quad 2x^2 - 5x + 2 \:=\:0 \quad\Rightarrow\quad(x - 2)(2x-1) \:=\:0 \)

. . Therefore: .\(\displaystyle x \:=\:2,\:\dfrac{1}{2}\)