How do you solve #11 + sqrt(2x+1)=2#?

It was suggested I post this

4 Answers
Jan 16, 2018

Answer:

No solutions.

Explanation:

Our goal is to isolate #x# so we can find its value.

#11+sqrt(2x+1)=2#

#rArr sqrt(2x+1)=-9#

We see that the equation has no solutions because the radical operator over #RR# always returns positive values. So the equation has no solution.

Jan 16, 2018

Answer:

The answer is no solution.

Explanation:

Your goal here is to isolate the variable.

First, start by subtracting 2 from both sides, as such:

#11-2+sqrt(2x+1)=2-2#

#9 +sqrt(2x+1) = 0#

Then subtract 9 from both sides:

#9-9 +sqrt(2x+1) = 0-9#

#sqrt(2x+1)=-9#

Now square both sides:

#(sqrt(2x+1))^2=(-9)^2#

#2x+1=81#

Then subtract 1 from both sides:

#2x+1-1=81-1#

#2x=80#

Finally, divided both sides by 2:

#(2x)/2=80/2#

#x=40#

Plug the answer into the original equation to check if it is correct:

#11+sqrt(2(40)+1)=2#

#11+9=2#

#20!=2#

When plugged in, we can see that #x=40# is not correct, so this problem has no solution.

Jan 16, 2018

Answer:

No Real Solution

Explanation:

Begin by subtracting #11# from both sides

#cancel(11color(red)(-11))+sqrt(2x+1)=2color(red)(-11)#

#sqrt(2x+1)=-9#

Square both sides to eliminate the radical on the left side

#sqrt(2x+1)^color(red)2=-9^color(red)2#

#2x+1=81#

Subtract #1# from both sides

#2x+cancel(1color(red)(-1))=81color(red)(-1)#

#2x=80#

Divide both side by #2#

#cancel2/cancelcolor(red)2x=80/color(red)2#

#x=40#

However if we verify this by substituting #40# for #x# in the original expression we'll find:

#11+sqrt(2(color(red)40)+1)=2#

#11+sqrt(81)=2#

#11+9=2#

#20!=2#

#:.# There are no real solutions

Jan 19, 2018

Answer:

No solution for principle root #->+sqrt(2x+1)+11=2#

The convention is; that unless the question states
#+-sqrt("something")# we only use the #+sqrt("smething")# which is the principle root.

Explanation:

However, just for fun; lets investigate the scenario of #(-2)xx(-2)=+4=(+2)xx(+2)#

So #sqrt(4)=+-2" is definitely viable "ul("only if you ignore the convention")#

The convention states :# sqrt(4)" can only "=+2" but you need"+-sqrt(4)" to have "+-2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given: #sqrt(2x+1)+11=2#

Write as #+-sqrt(2x+1)+11=2 color(red)(larr" not the actual question")#

Subtract 11 from both sides

#+-sqrt(2x+1)=-9#

Clearly #+sqrt(2x+1)=-9# does not have a solution for #x in RR# as each side of the equals is the opposite sign.

However #-sqrt(2x+1)=-9# may have a solution as both sides are negative. Accepting this should not be done as it is against the convention.#color(red)(" Note that this is not the question as given")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Just for fun consider "-sqrt(2x+1)=-9)#

Square both sides

#2x+1 = +81#

#2x=80#

#x=40#
Which is not the the answer for the given question as written.

Tony B

The graph clearly shows no solution for #+sqrt(2x+1)=-9# which is the answer to the given question as written.

If the question had been #+-sqrt(2x+1)# then we could legitimately have stated the answer as being: #(x,y)->(40,9)# is a solution for #-sqrt(2x+1)=-9#