College Preparatory Mathematics

\begin{align*} x^4-13x^2+36=0 \amp \quad \text{Quadratic form, one exponent, $4$, double the other, $2$} \\ y=x^2 \amp\quad \text{New variable equal to the variable with smaller exponent} \\ y^2=x^4 \amp\quad \text{Square both sides} \\ y^2-13y+36=0 \amp\quad \text{Substitute} y \text{for} x^2 \text{ and} y^2 \text{for} x^4 \\ (y-9)(y-4)=0 \amp\quad \text{Solve. We can solve this equation by factoring} \\ y-9=0 \text{ or } y-4=0 \amp\quad \text{Set each factor equal to zero } \\ \underline{+9 +9} \qquad\underline{+4 +4} \amp\quad \text{Solutions for $y$ , need $x$. We will use} y = x^2 \text{equation} \\ y=9 \text{ or } y=4 \amp\quad \text{Solutions for $y$, need $x$. We will use } y=x^2 \text{equation} \\ 9=x^2 \text{ or } 4=x^2 \amp\quad \text{Substitute values for $y$} \\ \pm \sqrt{9}=\sqrt{x^2} \text{ or } \pm \sqrt{4}=\sqrt{x^2} \amp\quad \text{Solve using the even root property, simplify roots} \\ x= \pm 3, \pm 2 \amp\quad \text{Our Solutions} \end{align*}

\begin{align*} a^{-2}-a^{-1}-6=0 \amp \quad \text{Quadratic form, one exponent, $- 2$, is double the other, $- 1$} \\ b=a^{-1} \amp\quad \text{Make a new variable equal to the variable with lowest exponent} \\ b^2=a^{-2} \amp\quad \text{Square both sides} \\ b^2-b-6=0 \amp \quad \text{Substitute } b^2 \text{ for } a^{-2} \text{ and } b \text{ for } a^{-1} \\ (b-3)(b+2)=0 \amp\quad \text{Solve. We will solve by factoring} \\ b-3=0 \text{ or } b+2=0 \amp\quad \text{Set each factor equal to $0$} \\ \underline{+3 +3} \qquad \underline{-2 -2} \amp \quad \text{Solve each equation} \\ b=3 \text{ or } b=-2 \amp\quad \text{Solutions for $b$, still need $a$, substitute into } b = a^{-1} \\ 3=a^{-1} \text{ or } -2=a^{-1} \amp\quad \text{Raise both sides to $- 1$ power} \\ 3^{-1}=a \text{ or } (-2)^{-1} \amp \quad \text{Simplify negative exponents} \\ a= \frac{1}{3}, - \frac{1}{2} \amp\quad \text{Our Solution} \end{align*}

\begin{align*} 2x^4+x^2=6 \amp\quad \text{Make equation equal to zero} \\ \underline{-6 -6} \amp\quad \text{Subtract $6$ from both sides} \\ 2x^4+x^2-6=0 \amp\quad \text{Quadratic form, one exponent, $4$, double the other, $2$ } \\ y=x^2 \amp\quad \text{New variable equal variable with smallest exponent} \\ y^2=x^4 \amp\quad \text{Square both sides} \\ 2y^2+y-6=0 \amp\quad \text{Solve. We will factor this equation} \\ (2y-3)(y+2)=0 \amp\quad \text{Set each factor equal to zero} \\ 2y-3=0 \text{ or } y+2=0 \amp\quad \text{Solve each equation} \\ \underline{+3 +3}\qquad \underline{-2 -2} \\ \frac{2y}{2}=\frac{3}{2} \text{ or } y=-2 \\ y=\frac{3}{2} \text{ or } y=-2 \amp\quad \text{We have $y$, still need $x$. Substitute into } y=x^2 \\ \frac{3}{2}=x^2 \text{ or } -2=x^2 \amp\quad \text{Square root of each side} \\ \pm \sqrt{\frac{3}{2}}=\sqrt{x^2} \text{ or } \pm \sqrt{-2}=\sqrt{x^2} \amp\quad \text{Simplify each root, rationalize denominator} \\ x=\frac{\pm\sqrt{6}}{2}, \pm i\sqrt{2} \amp\quad \text{Our Solution} \end{align*}

\begin{align*} 3(x-7)^2-2(x-7)+5=0 \amp\quad \text{Quadratic Form} \\ y=x-7 \amp\quad \text{Define new variable} \\ y^2=(x-7)^2 \amp\quad \text{Square both sides} \\ 3y^2-2y+5=0 \amp\quad \text{Substitute values into original}\\ (3y-5)(y+1)=0 \amp\quad \text{Factor} \\ 3y-5=0 \text{ or } y+1=0 \amp\quad \text{Set each factor equal to zero} \\ \underline{+5 +5} \qquad\underline{-1 -1} \amp\quad \text{Solve each equation} \\ 3y=5 \text{ or } y=-1 \\ \frac{3y}{3}=\frac{5}{3} \text{ or } y=-1 \\ y=\frac{5}{3} \text{ or } y=-1 \amp\quad \text{We have $y$, we still need $x$} \\ \frac{5}{3}=x-7 \text{ or } -1=x-7 \amp\quad \text{Substitute into } y=x-7 \\ \underline{+\frac{21}{3} +7}\qquad\quad \underline{+7 \quad +7} \amp\quad \text{Add $7$. Use common denominator as needed} \\ x=\frac{26}{3}, 6 \amp\quad \text{Our Solution} \end{align*}

\begin{align*} (x^2-6x)^2=7(x^2-6x)-12 \amp\quad \text{Make equation equal zero} \\ -7(x^2-6x)+12-7(x^2-6x)+12 \amp\quad \text{Move all terms to left} \\ (x^2-6x)-7(x^2-6x)+12=0 \amp\quad \text{Quadratic form} \\ y=x^2-6x \amp\quad \text{Make new variable} \\ y^2=(x^2-6x)^2 \amp\quad \text{Square both sides} \\ y^2-7y+12=0 \amp\quad \text{Substitute into original equation} \\ (y-3)(y-4)=0 \amp\quad \text{Solve by factoring} \\ y-3=0 \text{ or } y-4=0 \amp\quad \text{ Set each factor equal to zero} \\ \underline{+3 +3} \qquad \underline{+4 +4} \amp\quad \text{ Solve each equation} \\ y=3 \text{ or } y=4 \amp\quad \text{We have $y$, still need $x$.} \\ 3=x^2-6x \text{ or } 4=x^3-6 \amp\quad \text{Solve each equation, complete the square} \\ (\frac{1}{2}\cdot 6)^2=3^2=9 \amp\quad \text{Add $9$ to both sides of each equation} \\ 12=x^2-6x+9 \text{ or } 13=x^2-6x+9 \amp\quad \text{Factor} \\ 12=(x-3)^2 \text{ or } 13=(x-3)^2 \amp\quad \text{Use even root property} \\ \pm\sqrt{12}=\sqrt{(x-3)^2} \text{ or } \pm\sqrt{13}=\sqrt{ (x-3)^2} \amp\quad \text{Simplify roots} \\ \pm 2\sqrt{3}=x-3 \text{ or } \pm \sqrt{13}=x-3 \amp\quad \text{Add $3$ to both sides} \\ \underline{+3 \qquad +3} \qquad \underline{+3 \qquad +3} \amp\quad \\ x=3 \pm 2\sqrt{3}, \ 3 \pm \sqrt{13} \amp\quad \text{Our Solution} \end{align*}

\begin{align*} x^6-9x^3+8=0 \amp\quad \text{Quadratic form, one exponent, $6$, double the other, $3$} \\ y=x^3 \amp\quad \text{New variable equal to variable with lowest exponent} \\ y^2=x^6 \amp\quad \text{Square both sides} \\ y^2-9y+8=0 \amp\quad \text{Substitute $y^2$ for $x^6$ and $y$ for } x^3 \\ (y-1)(y-8)=0 \amp\quad \text{Solve. We will solve by factoring. } \\ y-1=0 \text{ or } y-8=0 \amp\quad \text{Set each factor equal to zero} \\ \underline{+1 +1} \qquad \underline{+8 +8} \amp\quad \text{Solve each equation} \\ y=1 \text{ or } y=8\amp\quad \text{Solutions for $y$ , we need $x$. Substitute into} y = x^3 \\ x^3=1 \text{ or } x^3=8 \amp\quad \text{Set each equation equal to zero} \\ \underline{-1 -1}\quad \underline{-8 -8} \\ x^3-1=0 \text{ or } x^3-8=0 \amp\quad \text{Factor each equation, difference of cubes} \\ (x-1)(x^2+x+1)=0 \amp\quad \text{First equation factored. Set each factor equal to zero} \\ x-1=0 \text{ or } x^2+x+1=0 \amp\quad \text{First equation is easy to solve} \\ \underline{+1 +1}\qquad \qquad \qquad \quad \amp\quad \\ x=1 \qquad \qquad \qquad \quad \amp\quad \text{First solution} \\ \frac{-1 \pm \sqrt{1^2-4(1)(1)}}{2}=\frac{1 \pm i\sqrt{3}}{2} \amp\quad \text{Quadratic formula on second factor} \\ (x-2)(x^2+2x+4)=0 \amp\quad \text{Factor the second difference of cubes} \\ x-2=0 \text{ or } x^2+2x+4=0 \amp\quad \text{Set each factor equal to zero.} \\ \underline{+2 +2}\qquad \qquad \qquad \qquad \amp\quad \text{First equation is easy to solve} \\ x=2 \qquad \qquad \qquad \qquad \amp\quad \text{Our fourth solution} \\ \frac{-2 \pm \sqrt{2^2-4(1)(4)}}{2(1)} =-1 \pm i\sqrt{3} \amp\quad \text{Quadratic formula on second factor} \\ x=1, 2, \frac{1 \pm i\sqrt{3}}{2}, -1 \pm i\sqrt{3} \amp\quad \text{Our final six solutions} \end{align*}