Question

Question #e1353

Answer 1

`x=\pm\sqrt(14)`

Explanation:

Equation given

`x^2-2=12`

Method 1

Isolate the `x^2` and solve by square root.
`x^2\cancel(-2)\cancel(\color(blue)(+2))=12\color(blue)(+2)`
`x^2=14`
`x=\pm\sqrt(14)`

Method 2

Form a quadratic equation by making one side equal to 0, then solve by factoring said quadratic.
`x^2-2\color(red)(-12)=\cancel(12)\cancel(\color(red)(-12))`
`x^2-14=0` `lArr` the factor of variable `x` is 0x, so...
`\rArrx^2+0x-14=0` apply quadratic formula

quadratic formula below
`x=(-0\pm\sqrt(0^2-4(1)(-14)))/(2(1))=(\pm\sqrt(56))/2=(\pm2\sqrt(14))/2`
`\rArr(\pm\cancel(2)\sqrt(14))/\cancel(2)=\pm\sqrt(14)`