Question

No idea how to solve this: 4(rootsign)x+1-x=4

Answer 1

This?
\[4\sqrt{x+1} -x = 4\]

Answer 2

yes

Answer 3

\[4\sqrt{x+1} -x = 4\]
\[\implies 4\sqrt{x+1} = 4 + x\]
\[\implies 16(x+1) = (4+x)^2\]
Can you take it from there?

Answer 4

yes , u square on both sides and u'll get 2 roots.
one of them is called an extreneous root.
this occurs because, the given equation is linear(highest power of x is 1) and when u square both sides , u create a root( highest power now is 2).
note: to eliminate this exrteneous root, substitute the value back in the original equation to see if its valid.

solution for this equation is: x=8.

Answer 5

I don't think the other root is extraneous. x=0 is valid.

Answer 6

I think I can do this. I'm ultimately solving for x so by doing the distribution equation posted by polpak, I should get the answer of 0 and/or 8?

Answer 7

Yep.

Answer 8

question: when I see a number before the sqaureroot, do I also square that? Like in this one, is that how you got 16?

Answer 9

\[4\sqrt{x+1} -x = 4\]

Is not a linear equation.

Answer 10

yea i was wrong :/ sorry, 0 and 8 are the roots.

Answer 11

Yes, I squared both sides.

\[(4\sqrt{x+1})^2 = 4^2(\sqrt{x+1})^2 = 16(x+1)\]

Answer 12

Or more correctly \[16|x+1|\]

Answer 13

ok...lightbulb is atarting to come on :) I'm seeing it.

Answer 14

thank you!