How do you simplify #sqrt(2x+1 )= x-7#?

1 Answer
Apr 1, 2018

Answer:

#x = 12#

Explanation:

#sqrt(2x+1) = x-7#

First, we want to get rid of the square root, so we square both sides:

#sqrt(2x+1)^2=(x-7)^2#

which simplifies to...
#2x+1=x^2-14x+49#

Now we move everything to one side of the equation:
#0 = x^2-16x+48#

To solve this, we have to factor it.

To factor this, we have to see which two numbers:
1. Multiply up to #48#
2. Add up to #-16#.

When we come up with the factors of #48#, we will notice that #-4# and #-12# multiply up to #48# and add up to #-16#:
#-4*-12 = 48#
#-4 - 12 = -16#

So the factored form looks like this:
#0 = (x-4)(x-12)#

Now we can set:
#x-4 = 0# and #x - 12 = 0#
#x = 4# and #x = 12#


Now we need to test to make sure both of these numbers work in the original equation. So we just substitute one number at a time in place of each #x#.
Let's check #4# first:
#sqrt(2(4)+1) = 4-7#
#sqrt(9) = -3#
#3 != -3#

So we now that #4# is really not a solution.

Now let's check #12#:
#sqrt(2(12)+1) = 12-7#
#sqrt(25) = 5#
#5 = 5#
Yes, this solution works!

So #x = 12#.


By the way:
You have to check problems with square roots and fractions! Or else your answer may only be partially right or not right at all!

For other problems, if none of your solutions worked when you checked it, that means the answer is no solution or #cancelO#!

Hope this helps!