Question

solve √x+1+ 5=x. 5 is not in the square root

Answer 1

So the question is:
\[\sqrt{x + 1} + 5 = x\]

Answer 2

yes .

Answer 3

ive gotten so far as 0=x^2-x-26

Answer 4

x = 8

Answer 5

how did u get that ?

Answer 6

you can graph them and find the intersection point.

Answer 7

oh okay .

Answer 8

Step 1: Bring the 5 to the other side.
\[\sqrt{x + 1} = x - 5\]Step 2: Square both sides.
\[(\sqrt{x + 1})^2 = (x - 5)^2\]Step 3: FOIL
\[x + 1 = x^2 - 10x + 25\]Step 4: Set up a quadratic.
\[0 = x^2 - 11x + 24\]Step 5: Factor
\[0 = (x - 8)(x - 3)\]Step 6: Find the solutions
\[0 = x - 8\]\[x = 8\]\[0 = x - 3\]\[x = 3\]Step 7: Plug both solutions in to look for extraneous solutions. (In other words, plug these two in for x. If the final statement isn't true, it's not a real solution and is extraneous. Can you do step 7?

Answer 9

\[\sqrt{x+1} + 5 = x\]
\[\sqrt{x+1} = x - 5\]
\[(\sqrt{x+1})^2 = (x-5)^2\]
\[x + 1 = x^2 - 10x + 25\]
\[x^2 - 10x - x + 25 - 1 = 0\]
\[x^2 - 11x + 24 = 0\]
\[(x-8)(x-3) = 0\]
does that help?

Answer 10

or you can do it the long way , using what they did above lol

Answer 11

lol...it's important to know how to do these this way since it can take a while to graph it's exact points...

Answer 12

@Calcmathlete yes i can (:

Answer 13

Alright. As long as you know how to do it, that's all that matters :)

Answer 14

yes , u r very helpful ! thank you (: i know most people just want the answer but i really just want to see how to get to it .

Answer 15

Glad to help!

Answer 16

but, if you are allowed to use a graphing calculator, you can plot the right side and the left side as two functions, and then use the Find Intersections :) Casios are the best for that. :P (saves time)

Answer 17

tru , well ill look into that ! thanks (:

Answer 18

square root on both sides is easy method..