Question

need help to explain this correctly step by step

Answer 1

need help to explain this correctly step by step
justify each step, and explain how the solution is extraneous.

√x + 4 = x- 2
x = 0 Extraneous Solution no solutions

Answer 2

please I'm desperate need of help this just practice for my exam, i just want to be familiar on how to explain stuff like this

Answer 3

is it
\[ \sqrt{x+4}= x-2 \]
?

Answer 4

no the equation is solved i just need to explain properly how to solve and what makes it extraneous

Answer 5

i tried to explain it but apparently i was wrong

Answer 6

i want help because this is my review and i need to learn how to explain this for my final exam

Answer 7

What is the original problem?
√x + 4 = x- 2
means
\[ \sqrt{x} + 4 = x -2 \]

Answer 8

√x + 4 = x- 2
x = 0 Extraneous Solution no solutions

Answer 9

so only the x is inside the square root ?

Answer 10

idk i just need to explain step by step what you do

Answer 11

ik its extraneous because it has no solutions right

Answer 12

to solve the equation, you have to be clear what equation you are talking about.

which problem are you solving:
\[ \sqrt{x} + 4 = x -2 \]
or
\[ \sqrt{x+ 4} = x -2 \]

Answer 13

the second one

Answer 14

good. now before solving it, let's just test if x=0 works.
start with
\[ \sqrt{x+4} = x-2\]
replace x with 0, and simplify. what do you get ?

Answer 15

I'm sorry i don't completely get this stuff

Answer 16

you would add to both sides correct

Answer 17

we are going to "test" if x=0 works. everywhere you see x, erase it, and write 0 in its place
in
\[ \sqrt{x+4}= x - 2\]
can you do that ?

Answer 18

√0 + 4 = 0- 2

Answer 19

and what is 0+4 ? and 0-2 ?

Answer 20

4 and -2

Answer 21

so you now have
\[ \sqrt{4} = - 2 \]
when solving radical equations, you only take the POSITIVE square root
on the right side you get the positive square root of 4
or
2 = -2 which is not correct. That shows that x=0 is not a true solution.

got that part ?

Answer 22

yes that makes sense

Answer 23

now, let's "solve"
\[ \sqrt{x+4}= x- 2 \]
to "get rid of" the square root, square both sides of the equation
can you do that ? what do you get ?

Answer 24

16 and 4

Answer 25

so √x + 16 = x-4

Answer 26

if you just had a single number, you could do that. But "square" means "multiply by itself"

If you have a complicated thing, it is still "complicated thing" times "complicated thing"

to square both sides, you multiply each side by itself:
\[ ( \sqrt{x+4} )^2= (x- 2)^2 \\
\sqrt{x+4} \cdot \sqrt{x+4} = (x-2)(x-2)
\]

Answer 27

i see

Answer 28

on the left side, you should learn that square root times itself "undoes" the square root
example
\[ \sqrt{A+B}\cdot \sqrt{A+B} = A+B\]

Answer 29

what do you get after simplifying the left side ?

Answer 30

x+ 8 if thats the left side

Answer 31

almost. all you do is "drop" the square root sign

examples:
\[ \sqrt{4}\cdot \sqrt{4}= 4 \\
\sqrt{3x+y}\cdot \sqrt{3x+y}= 3x+y \\
\sqrt{z^2+77}\cdot \sqrt{z^2+77}= z^2+77
\]
and so on.

try again

Answer 32

lol I'm sorry where did the 3 come from

Answer 33

√4 . √4 = 8

Answer 34

\[\sqrt{x+4} \cdot \sqrt{x+4} = (x-2)(x-2) \]
can you simplify the left side ?

Answer 35

and no, √4 . √4 is not 8

Answer 36

16?

Answer 37

you can't ignore the x in \( \sqrt{x+4} \)
look back at my post up above, with the examples. Look for the pattern.

Answer 38

√x+ √x+4=(x−2)(x−2)

Answer 39

Look at these examples
\[\sqrt{4}\cdot \sqrt{4}= 4 \\
\sqrt{3x+y}\cdot \sqrt{3x+y}= 3x+y \\
\sqrt{z^2+77}\cdot \sqrt{z^2+77}= z^2+77
\]
can you use the same pattern to simplify
\[ \sqrt{x+4} \cdot \sqrt{x+4} \]?

Answer 40

no

Answer 41

I'm sorry i don't understand this part

Answer 42

the idea is that
\[ \sqrt{A} \cdot \sqrt{A} = A \]
when you multiply a square root by itself, you get rid of the square root
so if you multiply
\[ \sqrt{stuff}\cdot \sqrt{stuff} = stuff\]
can you use that idea on
\[ \sqrt{x+4} \cdot \sqrt{x+4} \]

Answer 43

yes

Answer 44

what do you get ?

Answer 45

√4 . √4 = 256 or √4 . √4 = 16