Question

Hopefully one last question:
square root of x+6 + square root of x = 2.
I worked it out and got 1/4. How do I go back and check this? I don't know how to figure the 1/4 fraction under the radical sign.

Answer 1

it seems very unlikely to me that \(x=\frac{1}{2}\)

Answer 2

if you compute
\[\sqrt{\frac{1}{2}+6}+\sqrt{\frac{1}{2}}\] i doubt you get 2

Answer 3

is this the question
\[\sqrt{x+6}+\sqrt{x}=2\]?

Answer 4

Yes, that is the question.

Answer 5

then it cannot have a real solution

Answer 6

since \(x\geq 0\) the very smallest the left hand side can be is \(\sqrt{6}\) which is larger than 2

Answer 7

I ended up with 2=4(square root of x). squared each of those and got 1/4. That's where I know 1/4 was the wrong answer.

Answer 8

Can you show me a step by step?

Answer 9

I moved the 2nd radical x to the other side to get 2 - radical 2.

Answer 10

sure but before we begin lets understand that this is a fools errand because there is no solution to this problem
the left hand side of the equal sign in greater than or equal to \(\sqrt{6}\) no matter what \(x\) is, and so it cannot be equal to \(2\)

Answer 11

you mean you got \(2-\sqrt{x}\) right?

Answer 12

yes. then I squared both sides

Answer 13

ok so you get what when you square both sides?

Answer 14

x+6 = 4 - 4(radical x) + x

Answer 15

because i am going to guess that this is where the mistake is

Answer 16

on no you are right!

Answer 17

i then put like terms together to get 2 = -4(radical x)

Answer 18

I switched them to make a positive 4(radical x) = -2

Answer 19

\[x+6=4-4\sqrt{x}+x\] or\[ 2=-4\sqrt{x}\] now stop before you go any further

Answer 20

i squared both sides to get 16x=4

Answer 21

no no stop and think

Answer 22

\[4\sqrt{x}=-2\] right?

Answer 23

yes

Answer 24

ok now \(-2\) is pretty clearly negative, isn't it?

Answer 25

yes

Answer 26

and \(4\sqrt{x}\) is pretty obviously positive isn't it?

Answer 27

yes

Answer 28

so there is no solution

Answer 29

well actually \(4\sqrt{x}\) could be zero if \(x=0\) but it still cannot be \(-2\)

Answer 30

ok! Thanks so much! If I ever get a fractional answer again, is there a way to check the answer by putting a fraction back into the radical?

Answer 31

squaring will only confuse things, because then the right hand side \(-2\) will become \(4\) which is positive, and so you will get a solution, but it will not be a solution to the original equation

Answer 32

scroll up and look at what i wrote at the very beginning
i replaces \(x\) by \(\frac{1}{4}\)
you can compute if you like but there is no need because it is clear that it doesn't solve the original problem

Answer 33

That is what I did, and I just didn't know how to check it once I knew the 1/4 was a wrong answer.

Answer 34

*replaced

Answer 35

well we can do it if you like
\[\sqrt{\frac{1}{4}}=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}\]

Answer 36

and \(\frac{1}{4}+6=\frac{25}{4}\) so
\[\sqrt{\frac{25}{4}}=\frac{5}{2}\]

Answer 37

then add and get
\[\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3\] which is not what you want, you want \(2\)

Answer 38

Thanks for taking the time to write that out for me. I see now where I made a mistake in my checking.

Answer 39

ok but before you go, don't forget you are a thinking person, not an equation solving machine
when you see \(\sqrt{x+6}\) you should realize that this is always greater than or equal to \(\sqrt{6}\) and similarly \(\sqrt{x}\geq 0\) meaning
\[\sqrt{x+6}+\sqrt{x}\geq \sqrt{6}\] and
\[\sqrt{6}>2\] so there is no solution from the start