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

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

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