Question

Need some help solving this radical equation:
^2√(x−5) = 2

Answer 1

\[\sqrt[2]{x-5} = 2^{2}\]

Answer 2

\[\sqrt[2]{x-5} = 2^{2}\]

Answer 3

@elite ?

Answer 4

@ParthKohli ?

Answer 5

@Safa102 @mayankdevnani

Answer 6

someone please help @mathstudent55 @Hayhayz @imqwerty @Luigi0210 @SolomonZelman

Answer 7

Solving for X?

Answer 8

yes! and identifying if it is an extraneous solution

Answer 9

Square it on both sides

Answer 10

that's where i got stuck because of the \[\sqrt[2]{x-5}\] i dont know how to square it with the index being 2 like that

Answer 11

It's still a square root lol so you can square it on both sides you will get
x -5 = 2^4 (i am pretty sure it's the same lol)

Answer 12

oh it's the same? so

x-5 = 4
+5 +5
x = 9?

Answer 13

but my options are:

x = 6, solution is not extraneous
x = 6, solution is extraneous
x = 11, solution is not extraneous
x = 11, solution is extraneous

Answer 14

2 to the power of 4 isn't 8 lol
It's 16
x-5 = 16
x = 11

Answer 15

I am not sure which one it is though. It could be D or C.

Answer 16

oh sorry i typed the equation wrong it reads

\[\sqrt[2]{x-5} = 2\]

Answer 17

i was looking at the second step i had written down and accidentally added 2^2

Answer 18

Ohhhh
x -5 = 4
x = 9

but that's not one of the options....hmmm

Answer 19

is the ^2 at the beginning multiplying the equation?

Answer 20

can you post a screen shot of the original equation ?

Answer 21

Well @phi is here you don't need me anymore lol.

Answer 22

thank you @Serenity74 for your help ! @phi here is the screenshot

Answer 23

Lousy font. I would think the first 2 was meant to show square root, but it is the same size and shape as the 2 on the other side. Also, we won't get an answer that matches the choices if we assume the 2 means square root. So they probably mean
\[ 2 \sqrt{x-5}= 2\]

Answer 24

Though you could first square both sides (to get rid of the square root sign),
I would first divide both sides by 2

Answer 25

Oh that is true! And alright, if I squared both sides would I still get the same answer if I divided?

Answer 26

yes, if you don't make a mistake
what do you get ?

Answer 27

\[\sqrt{x-5}\]

Answer 28

did i delete something? im really bad at using openstudy sorry

Answer 29

but i think i got just the squareroot x -5 =1

Answer 30

if you divide by 2 on both sides you would get
\[ \frac{2}{2} \sqrt{x-5}= \frac{2}{2} \]
and 2/2 is 1 so that becomes
\[ \sqrt{x-5}=1 \]
now square both sides

Answer 31

I assume you know 1*sqrt(x-5) is just sqrt(x-5) because
1 times anything is the anything

Answer 32

once you get
\[ \sqrt{x-5}= 1\]
squaring the left side gets rid of the radical
on the right 1*1 is 1

Answer 33

yes! I finished the rest of the equation and got x=6. Can I ask, why does nothing happen to the radical when you divide 2 on both sides?

Answer 34

think of the radical (for the moment) as just some number (pretend it's 3)
then you are asking what happens when we divide
\[ 2 \cdot 3 \]
by 2
dividing by 2 is the same as multiplying by ½
so it would be
\[ \frac{1}{2} \cdot 2 \cdot 3 \]
when you multiply you can do it in any order. we can do the ½ * 2 first
\[ \left( \frac{1}{2}\cdot 2\right) \cdot 3 \]
and ½ * 2 is 1
\[ 1 \cdot 3\]
or just 3

the same thing happens if 3 were sqrt(x-5)

Answer 35

notice that is different from *adding*
for example 2+3 divided by 2 is not so nice:
\[ \frac{2+3}{2} = \frac{2}{2} + \frac{3}{2} \]
and you see we have to divide both terms by 2.

Answer 36

back to your problem. check that 6 "works" in the original equation. If it does, it is *not* an extraneous solution.

in this case
2*sqrt(6-5) = 2
2*sqrt(1) = 2
2*1 = 2
2=2
it works.

Answer 37

thank you for explaining, that makes perfect sense now! and i was just about to ask about the extraneous thing. I thought you had to multiply 2 by 6 and 5, then subtract. And does multiplying the square root by 2 cancel out the radical symbol?

Answer 38

do you remember order of operations? PEDMAS (or whatever acronym you might have seen)
the square root is treated like an exponent (½)
so to evaluate
\[ 2 \sqrt{6-5} \]
think of it as
\[ 2 \cdot (6-5)^\frac{1}{2} \]
first do the stuff in parens
then do the exponents
then the multiply

***And does multiplying the square root by 2 cancel out the radical symbol?***
no, only "squaring" cancels out a square root

Answer 39

so the first thing you do is 6-5 and get 1
\[ 2 \cdot 1^\frac{1}{2} \]
1 to *any power* is still 1 (so we can do that 1^½ in our head... we get 1)
\[ 2 \cdot 1 \]
finally 2 times 1 is 2

Answer 40

oh okay! thank you so much for your help, you're a lifesaver. My school doesn't do the best job at explaining mathematics. I do have two more to go and would really appreciate if you would help :) The first one is just some help on checking to see if the solution is extraneous and the second is a "see where eloise went wrong" type question.

Answer 41

please make a new post.

Answer 42

will do! :)