# Algebra Equations Reducible to Quadratics

## Question 1

Solve $x^4-5x^2 + 4 = 0$.

\begin{align} (x^2-4)(x^2-1) &= 0 \\ x^2-4 &= 0 \text{ or } x^2-1 = 0 \\ x^2 &= 4 \text{ or } x^2 = 1 \\ \therefore x &= \pm 2 \text{ or } x = \pm 1 \end{align}

## Question 2

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

\begin{align} \displaystyle x \color{green}{\times x} + \frac{2}{x} \color{green}{\times x} &= 3 \color{green}{\times x} \\ x^2 + 2 &= 3x \\ x^2-3x + 2 &= 0 \\ (x-1)(x-2) &= 0 \\ x-1 &= 0 \text{ or } x-2 = 0 \\ \therefore x &= 1 \text{ or } x = 2 \end{align}

## Question 3

Solve $(x+2)^2-3(x+2)-4 = 0$ by letting $A = x+2$.

\begin{align} \require{AMSsymbols} A^2-3A-4 &= 0 \\ (A-4)(A+1) &= 0 \\ A-4 &= 0 \text{ or } A+1 = 0 \\ A &= 4 \text{ or } A = -1 \\ x+2 &= 4 \text{ or } x+2 = -1 \\ \therefore x &= 2 \text{ or } x = -3 \end{align}

## Question 4

Solve $(x^2-x)^2-18(x^2-x) + 72 = 0$ by letting $A = x^2-x$.

\begin{align} A^2-18A + 72 &= 0 \\ (A-6)(A-12) &= 0 \\ A-6 &= 0 \text{ or } A-12 = 0 \\ A &= 6 \text{ or } A = 12 \\ x^2-x &= 6 \text{ or } x^2-x = 12 \\ x^2-x-6 &= 0 \text{ or } x^2-x-12 = 0 \\ (x-3)(x+2) &= 0 \text{ or } (x-4)(x+3) = 0 \\ \therefore x &= 3,-2,4,-3 \end{align}