Question

What is the solution to: √2x+13 -5=x?

A. x = -13
B. x = -10/11
C. x = 6
D. x = 3/7

Answer 1

is it \[\sqrt{2x+13-5}=x\text{ or }\sqrt{2x+13}-5=x\text{ or }\sqrt{2x}+13-5=x\]

Answer 2

the 3rd option, ^^^^^^^

Answer 3

in any of the cases, you need to isolate the square root on one side of the equation and then square both sides to remove the square root.

when you solve, you may get extraneous solutions so you need to check each solution in the original equation to ensure it is valid.

Answer 4

does that help?

Answer 5

not partiularly, still not sure

Answer 6

I think it's the second option, but I trust you :)

Answer 7

So here's an example: solve \(\sqrt{8x}-3+1=x\).
First we put everything except the radical on the right hand side:\[\sqrt{8x}=x+2\]
Then we square both sides:\[8x=(x+2)^2\]
Then we expand the right hand side:\[8x=x^2+4x+4\]
Then form a quadratic equation:\[x^2-4x+4=0\]
Then solve the equation (I assume that you know how to solve a quadratic equation):\[x=2 (repeated)\]

Answer 8

does that help? :)

Answer 9

\[\sqrt{2x}+13-5=\sqrt{2x}+8=x \Rightarrow \sqrt{2x}=x-8\]
\[\Rightarrow \left( \sqrt{2x} \right)^2=(x-8)^2\Rightarrow 2x=x^2-16x+64\]
\[\Rightarrow x^2-18x+64=0\]
Solve for x using whatever method you choose. Check your solutions in the original equation to make sure they work.

Answer 10

is it \[\sqrt{2}x+13-5=x?\]

Answer 11

you're problem needs to be stated clearly. I don't know what all is under the square root.

Answer 12

yes, it is the one you just said 2 comments ago.

Answer 13

so the x is not under the square root?

Answer 14

the x should be under the square root or else the answer would be irrational