Question

Please help me out real quick.​

Answer 1

`\text{B. }3x+4=3x+2√(3x-11)-10` only for
`3x+4\in [0, \infty)`

Explanation:

Key formulas used:

• 
  `√(x)^2=|x|`
• 
  `(a+b)^2=a^2+2ab+c^2`

Given
`√(3x+4)=1+√(3x-11)`,

We can use the first formula on the left side of the equation.

In this case, for
`√(x)^2=|x|`,
`x=3x+4`, we have:

`√(3x+4)^2=|3x+4|`

Similarly, we can use the second formula on the right side of the equation.

In this case, for
`(a+b)^2=a^2+2ab+c^2`,
`a=1, b=√(3x-11)`, we have:

`(1+√(3x-11))^2=1^2+2\cdot 1\cdot √(3x-11)+√(3x-11)^2,\\=1+2√(3x-11)+3x-11,\\=3x+2√(3x-11)-10`

Therefore, when you square both sides of the equation, you get:

`\boxed{\text{B. }3x+4=3x+2√(3x-11)-10}`

*Important:

This answer choice is actually only correct if
`3x+4>0`, because of the first formula we used. If
`3x+4<0` (negative), then
`3x+4\\eq √(3x+4)^2,\text{ if }3x-4<0`. Graphically, you can show this since the line
`y=√(x)^2` is not equal to
`y=x` but instead
`y=|x|`.
`y=√(x^2)` and
`y=x` only overlap if you restrict the domain to
`[0, \infty)` (positive numbers), hence
`\text{B. }3x+4=3x+2√(3x-11)-10` only for
`3x+4\in [0, \infty)`.