# Fraction Equations Reducible to Quadratic

\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} \\ 