Question

sqrt (x+1) +3=sqrt 3x+4
The +3 is on the outside of the sqrt.

Answer 1

square both sides

Answer 2

move the 3 to the right first

Answer 3

\[\sqrt{x + 1} + 3 = \sqrt{3x + 4} \]

Answer 4

right?

Answer 5

is the3x + 4 all under the root?

Answer 6

x=15

Answer 7

I think you should square on both sides.

Answer 8

can you explain?

Answer 9

assuming, that pratu's is the correct question... thats how I interpreted it as well.

Answer 10

please explain.

Answer 11

square both sides,
expand left,
collect likes,
factor,
square both sides,
expand right,
collect to the right,
complete the square,
solve for factors.

Is how I did it at least,

Answer 12

Pratu is correct. I don't get 15 even though that is the correct answer

Answer 13

\[(\sqrt{x+1}+3)^2=(\sqrt{3x+4}^2\]
\[x+1+6\sqrt{x+1}+9=3x+4\]
\[6\sqrt{x+1}=2x-6\]
\[36x+36=4x^2-24x+36\]
\[4x^2-60x=0\]
\[4x(x-15)=0\]
x=15

Answer 14

thanks.

Answer 15

Thats not quite how I did it, but yeah... lol :D

Answer 16

Where does the (x+1+6) come from in this part of the problem?[x + 1 +6\sqrt{x}+9\] I understand the 9 but not sure where the 6 is from

Answer 17

\[(\sqrt{x+1})^2=x+1\] and\[2*3*\sqrt{x+1}=6\sqrt{x+1}\], because
\[(a+b)^2=a^2+2ab+b^2\]

Answer 18

\[\sqrt{x+1}+3 = \sqrt{3 x+4}\]
\[(\sqrt{x+1}+3)^2 = 3 x+4\]
\[x+6 \sqrt{x+1}+10 = 3 x+4\]
\[6 \sqrt{x+1} = 2 x-6\]
\[\sqrt{x+1} = \frac16 (2 x-6)\]
\[x+1 = \frac1{36} (2 x-6)^2\]
\[x+1 = \frac{x^2}{9}-\frac{2 x}3+1\]
\[\frac{5 x}3-\frac{x^2}9 = 0\]
\[x^2-15 x = 0\]
\[x^2-15 x+\frac{225}4 = \frac{225}4\]
\[(x-15/2)^2 = \frac{225}4\]
\[|x-15/2|= 15/2\]
x-15/2 = -15/2 or x-15/2 = 15/2
x = 0 or x-15/2 = 15/2
x = 0 or x = 15

Looking at the case for 0, you'll see it isnt a solution.

Answer 19

Thanks to everyone!