Equations Reducible to Quadratic by Substitution

Equations Reducible to Quadratic by Substitution

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

Question 1

Solve \( (x+2)^2-3(x+2)-4 = 0 \).

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

Question 2

Solve \( (x-1)^4-11(x-1)^2 + 18 = 0 \).

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

Question 3

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

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

Question 4

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

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

 

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