# Fraction Equations Reducible to Quadratic

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

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

## Practice Questions

### Question 1

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

\begin{aligned} \displaystyle \require{AMSsymbols} \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} \require{AMSsymbols} \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{AMSsymbols} \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{AMSsymbols} \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}