Question

the square root of 3x+1-the square root of x+1=2

Answer 1

\[\sqrt{3x+1}-\sqrt{x+1}=2\]

Answer 2

i know the answer is 1 but I got 2 and dont know how to get 1 except if i dont square the 2

Answer 3

hi are you stuck? I have been for 15 minutes on this lol should i just skip it?

Answer 4

\[\begin{align}
\sqrt{3x+1}-\sqrt{x+1}&=2\\
(\sqrt{3x+1}-\sqrt{x+1})^2&=4\\
(3x+1)-2\sqrt{(3x+1)(x+1)}+(x+1)&=4\\
-2\sqrt{3x^2+4x+1}&=-4x+2\\
4(3x^2+4x+1)&=(-4x+2)^2\\
12x^2+16x+4&=16x^2-16x+4\\
0&=4x^2-32x\\
0&=4x(x-8)
\end{align}\]
It took me so long because I was confused about the x=0 part. Doesn't seem like it works, but if you take the first \(\sqrt{1}\) as 1 and the second as -1, it works. And the x=8 definitely works.

Answer 5

wait what ? 3rd line?

Answer 6

i thought i was supposed to just move the radx+1 to the other side and square each side

Answer 7

\[\begin{align}
(\sqrt{3x+1}-\sqrt{x+1})^2&=4\\
(\sqrt{3x+1}-\sqrt{x+1})(\sqrt{3x+1}-\sqrt{x+1})&=4\\
(\sqrt{3x+1}\cdot\sqrt{3x+1})+(\sqrt{3x+1}\cdot-\sqrt{x+1})+\\(-\sqrt{x+1}\cdot\sqrt{3x+1})+(-\sqrt{x+1}\cdot-\sqrt{x+1})&=4\\
(3x+1)-2\sqrt{(3x+1)(x+1)}+(x+1)&=4\\
\end{align}\]

Answer 8

Wow thanks so much.

Answer 9

There might be an easier way to do it, that's just what looked easiest to me.

Answer 10

other way :
Put \[a=\sqrt {3x+1}\] and \[b=\sqrt{x+1}\] ,\[ a,b\ge 0\]
We have system of equation :
first equation is \[a^2-3b^2=-2\]
second is \[a-b=2\]
We can use substitution method here.