Question

Solve each equation by taking square roots 1. 9x^2-36=0 2. (x-3)^2+8=9 3. 5x^2-80 = 0 4. (x-12)^2 = 16 5. (2x - 1)^2-7=18 6. (x+8)² = 25 7. (x-5)^2 -6=13 8. 3x^2 +12=09. 3(x+1)^2-3=9010. 3x^2+4=-2311. 3x^2+10=5512.6x^2+5=143

Answer 1

`\begin{gathered} 1)9x^2=36 \\ x^2=(36)/(9) \\ x^2=4 \\ x=\sqrt[]{4} \\ x=|2| \end{gathered}`
`\begin{gathered} 2)(x-3)^2+8=9 \\ (x-3)^2=9-8 \\ (x-3)^2=1 \\ (x-3)^{}=|1| \\ \\ We\text{ have 2 cases, when } \\ x-3=1\text{ and x-3=-1} \\ IN\text{ THE FIRST CASE} \\ x=1+3 \\ x=4 \\ In\text{ the second case} \\ x-3=-1 \\ x=-1+3 \\ x=2 \\ \text{hence, the solutions are} \\ x=4\text{ and x=2} \end{gathered}`
`\begin{gathered} 3)5x^2-80=0 \\ 5x^2=80 \\ x^2=(80)/(5) \\ x^2=16 \\ x=\sqrt[]{16} \\ x=|4| \end{gathered}`
`\begin{gathered} 4)(x-12)^2=16 \\ (x-12)=\sqrt[]{16} \\ (x-12)=|4| \\ \text{the, we have 2 cases:} \\ (x-12)=4 \\ \text{and } \\ (x-12)=-4 \\ In\text{ the first case:} \\ x=4+12\Rightarrow x=16 \\ In\text{ the second case} \\ x=-4+12\Rightarrow x=8 \\ \text{therefore, the answers are} \\ x=16\text{ and x=8} \end{gathered}`
`\begin{gathered} 5)(2x-1)^2-7=18 \\ (2x-1)^2=18+7 \\ (2x-1)^2=25 \\ (2x-1)=\sqrt[]{25} \\ (2x-1)=|5| \\ \text{hence, we have 2 cases:} \\ \text{when }(2x-1)=5 \\ \text{and }(2x-1)=-5 \\ IN\text{ the first case:} \\ 2x=5+1 \\ 2x=6\Rightarrow x=3 \\ IN\text{ the swcond case} \\ 2x=-5+1 \\ 2x=-4\Rightarrow x=(-4)/(2)\Rightarrow x=-2 \\ \text{the solutions are} \\ x=3\text{ and x=-2} \end{gathered}`
`\begin{gathered} 6)(x+8)^2=25 \\ (x+8)=\sqrt[]{25} \\ x+8=|5| \\ \text{again, we have 2 cases:} \\ x+8=5 \\ \text{and} \\ x+8=-5 \\ IN\text{ the first case} \\ x=5-8\Rightarrow x=-3 \\ IN\text{ the second case} \\ x=-5-8\Rightarrow x=-13 \\ \text{hence, the solutions are} \\ x=-3\text{ and x=-13} \end{gathered}`