Question

Square root 4x+1 - square root x-2=3

Answer 1

\[\sqrt{4x+1}-\sqrt{x-2}=3\]
\[\sqrt{41+1}=3+\sqrt{x-2}\] now you are going to have to square twice

Answer 2

the first time you get
\[4x+1=9+x-2+6\sqrt{x-2}\]

Answer 3

or
\[4x+1=7+x+6\sqrt{x-2}\] now isolate the radical

Answer 4

how did you get that?

Answer 5

okay lets go slow

Answer 6

start here
\[\sqrt{4x+1}=3+\sqrt{x-2}\] and square both sides to get rid of the radical
we get
\[(\sqrt{4x+1})^2=\left(3+\sqrt{x-2}\right)^2\]

Answer 7

the left hand side is \(4x+1\) because you are squaring the square root

Answer 8

can't you leave the √x-2 on the left side?

Answer 9

you can, but it will make a big big mess if you do

Answer 10

oh okay

Answer 11

once we have
\[(\sqrt{4x+1})^2=\left(3+\sqrt{x-2}\right)^2\] we get
\[4x+1=(3+\sqrt{x-2})(3+(\sqrt{x-2})\] and you have to multiply out

Answer 12

multiply out on the right using the distributive law gives
\[4x+1=3\times 3+3\times \sqrt{x-2}+3\times \sqrt{x-2}+\sqrt{x-2}\times \sqrt{x-2}\]
or
\[4x+1=9+6\sqrt{x-2}+x-2\]

Answer 13

so one radical is gone, but one still remains
now i combine like terms on the left, then isolate the radical so you can square again

Answer 14

\[4x+1=7+x+6\sqrt{x-2}\]
\[3x-6=6\sqrt{x-2}\]

Answer 15

now you have to square one more time to get rid of the radical on the right
don't forget to square the six too
you get
\[(3x-6)^2=36(x-2)\] and now you have a quadratic equation to solve

Answer 16

4x+1=7+x+6√x-2
how did you get the 7? i thought it was 6

Answer 17

fortunately this one is not too hard to solve
\[9x^2-36x+36=36x-72\]
\[9x^2-72x+108=0\] divide by 9

Answer 18

did you understand how i got this line?
\[4x+1=9+6\sqrt{x-2}+x-2\]

Answer 19

the \(7\) came from the \(9-2\)

Answer 20

ohh!! okay

Answer 21

sorry i was a little lost on that one line, yeah i understand (:

Answer 22

so you'd get x^2-8x+12?

Answer 23

yes, now factor and solve and check your answers