Question

Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.) 2x + the square root of x + 13 = 10

Answer 1

Start by moving the square root of x to one side of = n everything else to the other side...

Answer 2

Okay I will help by moving...\[\sqrt{x} = 2x...\]
then what?

Answer 3

okay i got square root of x+13=10-2x

Answer 4

Move the number to the left side plz? :)

Answer 5

too which then I assume we must get ride of the square root so we square the entire thing giving us (square root of x=13)squared=(10-2x)squared or what is it that you are suggesting?

Answer 6

:) u hardly need my help at all - yup dats da plan!

Answer 7

oh wait wait u need to move 13 over the right first though

Answer 8

(square root of x+13)squared my bad but then we get x+13=(10-2x)(10-2x) which then we foil the 10-2x side right?

Answer 9

(square root of x +13) squared will give x + 2x13x square root of x so that is no good. U should move 13 over the right before squaring,

Answer 10

so it's Square root of x=(10-2x-13)squared right?

Answer 11

Simplify 10-2x-13 first...then square both side of the equation...

Answer 12

\[(\sqrt{x+13})^{2}=(10-2x)^{2}\]

Answer 13

Oh sorry I mis-read ur Q - didnt know the 13 is within the SQRT. So yes u are on the right track :)

Answer 14

sorry first time using this place. So after all this we foil the (10-2x)(10-2x) we get \[x+13=100-20x-20x -4x^{2} \] which we combine the 20xs into 40x so we have once more \[x+13=\frac{ 100-40x+4x ^{2} }{ 4 }\]

Answer 15

Now u need to move everything to one side n combine n simplify to a standard quadratic eqn: ax^2 + bx + C = 0 to solve.

Answer 16

U are doing fine - sorry for my late replyl I was away...

Answer 17

thanks man for you're help

Answer 18

welcome u have helped urself the most :)