Question

Please help me out real quick.​

Answer 1

Answer:

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

Step-by-step explanation:

Key formulas used:

• $\sqrt{x}^2=|x|$
• $(a+b)^2=a^2+2ab+c^2$

Given $\sqrt{3x+4}=1+\sqrt{3x-11}$,

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

In this case, for $\sqrt{x}^2=|x|$, $x=3x+4$, we have:

$\sqrt{3x+4}^2=|3x+4|$

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

In this case, for $(a+b)^2=a^2+2ab+c^2$, $a=1, b=\sqrt{3x-11}$, we have:

$(1+\sqrt{3x-11})^2=1^2+2\cdot 1\cdot \sqrt{3x-11}+\sqrt{3x-11}^2,\\=1+2\sqrt{3x-11}+3x-11,\\=3x+2\sqrt{3x-11}-10$

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

$\boxed{\text{B. }3x+4=3x+2\sqrt{3x-11}-10}$

*Important:

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