\begin{aligned} \displaystyle \require{color} x+\frac{a}{x} &= b \\ x^2 + a &= bx \\ x^2 – bx + a &= 0 \\ \end{aligned}

Once, the equation forms a quadratic form by multiply the denominator to both sides, then the equation can be solved by quadratic solution, such as factorise or quadratic formula.
Let’s look at the following examples.

Practice Questions

Question 1

Solve $\displaystyle x+\frac{2}{x} = 3$.

\begin{aligned} \displaystyle \require{color} x^2 + 2 &= 3x &\color{red} \text{multiply both sides by } x \\ x^2 – 3x + 2 &= 0 \\ (x-1)(x-2) &= 0 \\ \therefore x = 1 &\text{ or } x=2 \\ \end{aligned} \\

Question 2

Solve $\displaystyle x^3 – \frac{8}{x^3} = 7$.

\begin{aligned} \displaystyle x^6 – 8 &= 7x^3 &\color{red} \text{multiply both sides by } x^3 \\ x^6 – 7x^3 – 8 &= 0 \\ (x^3+1)(x^3-8) &= 0 \\ x^3 = -1 &\text{ or } x^3 = 8 \\ \therefore x = -1 &\text{ or } x = 2 \\ \end{aligned} \\

Question 3

Solve $\displaystyle 16x^2 + \frac{16}{x^2} = 257$.

\begin{aligned} \displaystyle \require{color} 16x^4 + 16 &= 257x^2 &\color{red} \text{multiply both sides by } x^2 \\ 16x^4 – 257x^2 + 16 &= 0 \\ (16x^2 – 1)(x^2 – 16) &= 0 \\ (4x-1)(4x+1)(x-4)(x+4) &= 0 \\ \therefore x &= \frac{1}{4}, -\frac{1}{4}, 4, -4 \\ \end{aligned} \\

Question 4

Solve $\displaystyle \sqrt{x} + \frac{1}{\sqrt{x}} = 2$.

\begin{aligned} \displaystyle \require{color} \bigg(\sqrt{x} + \frac{1}{\sqrt{x}}\bigg)^2 &= 2^2 &\color{red} \text{square both sides} \\ x + 2 + \frac{1}{x} &= 4 \\ x – 2 + \frac{1}{x} &= 0 \\ x^2 – 2x + 1 &= 0 &\color{red} \text{multiply both sides by } x \\ (x-1)^2 &= 0 \\ \therefore x&= 1 \\ \end{aligned} \\