Answer:
[tex]\text{B. }3x+4=3x+2\sqrt{3x-11}-10[/tex] only for [tex]3x+4\in [0, \infty)[/tex]
Step-by-step explanation:
Key formulas used:
- [tex]\sqrt{x}^2=|x|[/tex]
- [tex](a+b)^2=a^2+2ab+c^2[/tex]
Given [tex]\sqrt{3x+4}=1+\sqrt{3x-11}[/tex],
We can use the first formula on the left side of the equation.
In this case, for [tex]\sqrt{x}^2=|x|[/tex], [tex]x=3x+4[/tex], we have:
[tex]\sqrt{3x+4}^2=|3x+4|[/tex]
Similarly, we can use the second formula on the right side of the equation.
In this case, for [tex](a+b)^2=a^2+2ab+c^2[/tex], [tex]a=1, b=\sqrt{3x-11}[/tex], we have:
[tex](1+\sqrt{3x-11})^2=1^2+2\cdot 1\cdot \sqrt{3x-11}+\sqrt{3x-11}^2,\\=1+2\sqrt{3x-11}+3x-11,\\=3x+2\sqrt{3x-11}-10[/tex]
Therefore, when you square both sides of the equation, you get:
[tex]\boxed{\text{B. }3x+4=3x+2\sqrt{3x-11}-10}[/tex]
*Important:
This answer choice is actually only correct if [tex]3x+4>0[/tex], because of the first formula we used. If [tex]3x+4<0[/tex] (negative), then [tex]3x+4\neq \sqrt{3x+4}^2,\text{ if }3x-4<0[/tex]. Graphically, you can show this since the line [tex]y=\sqrt{x}^2[/tex] is not equal to [tex]y=x[/tex] but instead [tex]y=|x|[/tex]. [tex]y=\sqrt{x^2}[/tex] and [tex]y=x[/tex] only overlap if you restrict the domain to [tex][0, \infty)[/tex] (positive numbers), hence [tex]\text{B. }3x+4=3x+2\sqrt{3x-11}-10[/tex] only for [tex]3x+4\in [0, \infty)[/tex].
