Question

Stuck on an algebra 2 question

Answer 1

sqrt(x - 8) + 3 = 1

Answer 2

Is the 3 under the square root?

Answer 3

move the 3 to the other side of the equation, then square the problem

Answer 4

I know that there is no solution, but I'm getting stuck here: sqrt(x - 8) + 3 = 1
sqrt(x - 8) = -2

I'm not sure if when I square both sides what the "-2" becomes.

No, the 3 isn't.

Answer 5

The -2 becomes a 4 when squared

Answer 6

Is this right?

sqrt(x - 8) + 3 = 1
sqrt(x - 8) = -2
x - 8 = 4
x = 4 + 8
x = 12

Answer 7

Yes that is :)

Answer 8

no because you only take the POSITIVE value when evaluating square roots

Answer 9

therefore it has no solution

Answer 10

Thank you, guys! Can you help me check it or whatever?

Answer 11

Yeah I know there is no solution, now I have to check it to show that it's extraneous.

Answer 12

come on only me insist that there is no solution doesn't mean that i'm wrong

Answer 13

It does have a solution.
sqrt (12-8)+3=1
sqrt(4)+3=1
2+3=1
5=1
It's extraneous

Answer 14

when you do square root, you only take the POSITIVE value

Answer 15

as zeinas said sqrt(12-8)+3=sqrt4+3=2+3=5 not 1

Answer 16

If x^2=4 then x can be +2 or -2 but sqrt4 is just +2

Answer 17

He's right, the x=12 is extraneous so it's not a solution.

Answer 18

KC is right.

Answer 19

https://en.wikipedia.org/wiki/Square_root
"Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root. For positive a, the principal square root can also be written in exponent notation, as \(a^{\frac12}\)."

Answer 20

Kc is right:
sqrt(x - 8) + 3 = 1
sqrt(x - 8) = -2
x - 8 = 4
x = 4 + 8
x = 12

Check

sqrt(12 - 8) + 3 = 1
12 - 8 + 3 = -1
4 + 3 = 1
7 = 1
The solution is extraneous.

Answer 21

Thanks for reminding me of the "rules". the square root of something can never be negative. Wolfram reinforces.

Answer 22

P.S I THINK I did that right haha

Answer 23

lol You should have done the 12-8 first, which is 4, then find the sqrt of 4 which is 2, then add that to three, but you were right about it being extraneous.

Answer 24

Oh okay. Can you guys explain one more thing to me? Please :) When you're checking it, how does it go from
= 1
= -1
= 1
= 1

Answer 25

Here would be my solution:

Consider the equation \(\sqrt{x-8}+3=1\).
\[\begin{array}{rcl}\sqrt{x-8}+3&=&1\\\sqrt{x-8}&=&1-3=-2\end{array}\]Since a square root of a number cannot be negative, it has no solution.

Answer 26

@HHS6281019 It's not supposed to change to -1. It should be +1 all the way. It was probably a typo.

Answer 27

Okay lol, that really confused me. Thanks again, guys! I'm not sure who to give a medal to?

Answer 28

give it to me

Answer 29

Check for x=12:
\[\begin{array}{}LHS&=\sqrt{12-8}+3\\&=\sqrt4+3\\&=2+3\\&=5\\\\RHS&=1\end{array}\]
Therefore x=12 is not a solution.

(P.S. I've already proved that it has no solution, so this step is unnecessary)

Answer 30

id say KC

Answer 31

:P

Answer 32

You don't have to give a medal. We all helped in a way

Answer 33

@zeinas. yep :D

Answer 34

psh i derserve that medal

Answer 35

here's your medal :)

Answer 36

lol kc u actually gave me one?

Answer 37

there ya go

Answer 38

lolz

Answer 39

i think there is a solution in complex numbers which is x =4i+8

Answer 40

remember \(\Large\sqrt i=\dfrac1{\sqrt2}(1+i)\)

Answer 41

Check:
sqrt(4i+8-8)+3
=sqrt(4i)+3
=2sqrt(i)+3
=2/sqrt(2)+2i/sqrt(2)+3
=sqrt(2)+sqrt(2)i+3

Answer 42

@mark_boltzman therefore x=4i+8 is not a solution :)

Answer 43

you r right i did a mistake :)

Answer 44

I also thought about complex number at the first place so it's fine :)

Answer 45

Wow that's some complicated stuff haha

Answer 46

lolz :P