# Equations Reducible to Quadratic by Substitution

Equations Reducible to Quadratic by Substitution to Learn Math is made the equations easy to solve by substitution to simplify the equations.

## Question 1

Solve $(x+2)^2 – 3(x+2) – 4 = 0$.

\begin{aligned} \displaystyle \require{color} \text{Let } A &= x+2 \\ A^2 – 3A -4 &= 0 &\color{red} \text{replace } x+2 \text{ by } A \\ (A-4)(A+1) &= 0 \\ A = 4 &\text{ or } A = -1 \\ x+2 = 4 &\text{ or } x+2 = -1 &\color{red} \text{replace } A \text{ by } x+2 \\ \therefore x = 2 &\text{ or } x = -3 \\ \end{aligned} \\

## Question 2

Solve $(x-1)^4 – 11(x-1)^2 + 18 = 0$.

\begin{aligned} \displaystyle \text{Let } A &= x-1 \\ A^4 – 11A + 18 &= 0 &\color{red} \text{replace } x-1 \text{ by } A \\ (A^2-9)(A^2-2) &= 0 \\ A^2 = 9 &\text{ or } A^2 = 2 \\ A = 3, -3 &\text{ or } A = \sqrt{2}, -\sqrt{2} \\ x-1 = 3, -3 &\text{ or } x-1 = \sqrt{2}, -\sqrt{2} &\color{red} \text{replace } A \text{ by } x-1 \\ \therefore x = 4, -2 &\text{ or } x = 1+\sqrt{2}, 1-\sqrt{2} \\ \end{aligned}

## Question 3

Solve $(x^2-x)^2 – 18(x^2-x) + 72 = 0$.

\begin{aligned} \displaystyle \text{Let } A &= x^2-x \\ A^2 – 18A + 72 &= 0 &\color{red} \text{replace } x^2-x \text{ by } A \\ (A-12)(A-6) &= 0 \\ A = 12 &\text{ or } A = 6 \\ x^2-x = 12 &\text{ or } x^2-x = 6 &\color{red} \text{replace } A \text{ by } x^2-x \\ x^2-x – 12 = 0 &\text{ or } x^2-x – 6 = 0 \\ (x-4)(x+3) = 0 &\text{ or } (x-3)(x+2) = 0 \\ \therefore x &= 4, -3, 3, -2 \\ \end{aligned}

## Question 4

Solve $\displaystyle \Big(x+\frac{1}{x}\Big)^2 + \Big(x+\frac{1}{x}\Big) – 2 = 0$.

\begin{aligned} \displaystyle \text{Let } A &= x+ \frac{1}{x} \\ A^2 + A – 2 &= 0 \\ (A+2)(A-1) &= 0 \\ A = -2 &\text{ or } A = 1 \\ x + \frac{1}{x} = -2 &\text{ or } x + \frac{1}{x} = 1 \\ x^2 + 1 = -2x &\text{ or } x^2 + 1 = -x \\ x^2 + 2x + 1 = 0 &\text{ or } x^2 +x + 1 = 0 \\ (x+1)^2 = 0 &\text{ or } x = \frac{1 \pm \sqrt{-3}}{2} &\color{red} \text{undefined}\\ \therefore x = -1 &\text{ } \end{aligned} \\