Question

Solve in square root 1: 9x^2-36=02: (x-3)^2+8=93: 5x^2-80=04: (x-12)^2=16

Answer 1

These are quadratic equations, and we need to find the roots, so:

`\begin{gathered} \text{Equation 1:} \\ 9x^2-36=0 \\ 9x^2=36 \\ x^2=(36)/(9)=4 \\ x=\pm\sqrt[]{4}=\pm2 \\ So,\text{ the solutions are x=2 and x=-2} \end{gathered}`
`\begin{gathered} \text{Equation 2:} \\ (x-3)^2+8=9 \\ (x-3)^2=9-8=1 \\ x-3=\pm\sqrt[]{1}=\pm1 \\ x=3\pm1,x_1=3+1=4,x_2=3-1=2 \\ So,\text{ the solutions are x=4 and x=2} \end{gathered}`
`\begin{gathered} \text{Equation 3}\colon \\ 5x^2-80=0 \\ 5x^2=80 \\ x^2=(80)/(5)=16 \\ x=\pm\sqrt[]{16}=\pm4 \\ So,\text{ the solutions are x=4 and x=-4} \end{gathered}`
`\begin{gathered} \text{Equation 4:} \\ (x-12)^2=16 \\ x-12=\pm\sqrt[]{16}=\pm4 \\ x=12\pm4,x_1=12+4=16,x_2=12-4=8 \\ So,\text{ the solutions are x=16 and x=8} \end{gathered}`