Please help me out, will medal!! I'm really confused. It's creating a radical equation. I do not understand radical equations at all.

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- horsegirl27

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- horsegirl27

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- horsegirl27

Could someone explain radical equations, then help me make 2 for this question? And what does extraneous mean?

- horsegirl27

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## More answers

- anonymous

1) Isolate the radical symbol on one side of the equation
2) Square both sides of the equation to eliminate the radical symbol
3) Solve the equation that comes out after the squaring process
4) Check your answers with the original equation to avoid extraneous values

- horsegirl27

How do I isolate it?

- phi

I think I would start by picking a number and squaring it.
for example
5
and 5*5= 25
and write
\[ \sqrt{25} = 5 \]

- horsegirl27

Okay.

- phi

they want
\[ a \sqrt{x+b} + c = d\]
so far we have
\[ \sqrt{25} = 5\]
we could write 25 as the sum of two numbers
can you pick a pair of numbers that add up to 25 ?

- horsegirl27

16 and 9?

- phi

ok. now let x be one of the numbers (which we pretend we don't know)
and b be the other number.
example: let x be 16
then we can write
\[ \sqrt{x+9}= 5 \]
(and we know x=16 would make this true)
now we want to multiply by "a"
\[ a\sqrt{x+9}= a\cdot 5 \]
we can pick any number we want for a
pick a number for a

- horsegirl27

3

- phi

If it is not clear, we are creating a complicated equation that matches what they want
\[ a \sqrt{x+b} + c = d \]

- horsegirl27

I understand how we're replacing numbers for the letters

- phi

so if we pick a to be 3, then
\[ a\sqrt{x+9}= a\cdot 5 \\ 3\sqrt{x+9}= 3\cdot 5 \\ 3\sqrt{x+9}=15\]

- phi

almost done. they want us to add a number "c" to the left side (and to keep things equal) we do the same to the right side
\[ 3\sqrt{x+9} + c=15+c \]

- horsegirl27

okay, should we use any number for c?

- phi

yes, we are making it up. It can be anything (but 0 would be cheating)

- horsegirl27

Okay let's use 7.

- phi

\[ 3\sqrt{x+9} + c=15+c \\ 3\sqrt{x+9} + 7=15+7 \\ 3\sqrt{x+9} + 7=22 \]

- phi

I don't know if this has an extraneous solution but I do know x=16 works because that is how we started... which means when you solve it , you better get x=16 as an answer

- horsegirl27

Do you know what extraneous means?

- phi

Extraneous means "extra and wrong"
to solve this equation we have to undo all the steps we used to make it.
to do that, we square the radical , and we get a complicated equation that has more than 1 solution (one of them we do not want)

- phi

Let's try to solve your equation
\[ 3\sqrt{x+9} + 7=22 \]
First add -7 to both sides

- horsegirl27

that will take out +7 and change 22 to 15

- horsegirl27

\[3\sqrt{x+9}=15 \]

- phi

now divide both sides by 3

- horsegirl27

how do I do that? I know 15 will become 5, but what about the left side?

- phi

I guess you have to know how fractions work. With just numbers:
\[ \frac{3 \cdot 2 }{3} = \frac{3}{3} \cdot 2 = 2 \]
If the 2 were something else (such as \( \sqrt{x+9} \) it still works the same way

- horsegirl27

ohhh ok

- phi

so you get
\[ \frac{3 \sqrt{x+9}}{3} = \frac{15}{3} \]

- phi

which simplifies to ?

- horsegirl27

Wait, after simplifying, if I divide 3 by 3, will that be changing it to 1?

- phi

on the left , the 3/3 = 1
on the right the 15/3 = 5

- horsegirl27

ok, I knew the right side

- phi

yes, 3/3 =1 and 1 times anything is the anything

- horsegirl27

So it will be \[1\sqrt{x+3}=5\] or is that wrong?

- phi

btw, notice
\[ \frac{15}{3} = \frac{3 \cdot 5}{3} = \frac{3}{3} \cdot 5 =1\cdot 5= 5 \]

- phi

You can only divide one thing. if you have
\[ \frac{3\cdot 3 \cdot 3}{3} = \frac{3}{3} \cdot 3 \cdot 3 \]
notice you don't divide the other 3's.
so in our problem the \(\sqrt{x+9} \) is not divided by 3
if that makes sense?

- horsegirl27

Ohhhh yes it finally makes sense now!

- phi

another way to keep it straight is to think of dividing as multiplying by the reciprocal
in other words
\[ \frac{3 \cdot 3 }{3} \text{ means } \frac{1}{3} \cdot 3 \cdot 3 \]
(change divide by 3 into multiply by 1/3)
it should be more clear you can do
\[ \left( \frac{1}{3} \cdot 3 \right) \cdot 3 \]

- horsegirl27

ok

- phi

and definitely not
\[ \left( \frac{1}{3} \cdot 3 \right) \cdot\left( \frac{1}{3} \cdot 3 \right) \]

- phi

but back to the problem. what do you have so far?

- horsegirl27

um I think \[1\sqrt{x+9}=5\]

- phi

yes, but generally we don't bother to show the multiply by 1

- horsegirl27

ok

- phi

in other words we know 1 times \(\sqrt{x+9}\) is \(\sqrt{x+9}\)
so we just write
\[ \sqrt{x+9}=5 \]

- horsegirl27

ok, that makes sense

- phi

we could write (goofy) things like
\[ \sqrt{x+9}= 1\cdot 5+ 0 \]
but we get into the habit of writing as few symbols as possible

- horsegirl27

Yeah, we don't need to write all of that

- phi

to "get rid" of the radical operator, square both sides
what do we get?

- horsegirl27

well, 9 squared is 81, and 5 squared is 25

- phi

you have
\[ \sqrt{x+9}=5 \]
"square both sides" means do
\[ \left(\sqrt{x+9}\right)^2 =5^2\]

- horsegirl27

ohhhhhhhhhhhhhhhhhhhhhhhh

- phi

square means "multiply by itself one time"
\[ \left(\sqrt{x+9}\right)\left(\sqrt{x+9}\right) =5 \cdot 5 \]

- horsegirl27

that makes sense now lol

- phi

by definition of square root
\[ \sqrt{z} \cdot \sqrt{z} = \sqrt{z\cdot z}= z \]
it does not matter how "complicated" z is (in our case (x+9) )

- horsegirl27

ok

- phi

or, just remember, squaring a square root undoes the square root

- phi

examples:
\[ \left( \sqrt{3x}\right)^2 = 3x \\
\left( \sqrt{x^2+y^2}\right)^2 = x^2 + y^2
\]

- horsegirl27

that sort of makes sense

- phi

It is a definition

- horsegirl27

Okay, reading it a few times, I understand

- horsegirl27

You square a square root so it's no longer a square root.

- phi

yes

- horsegirl27

ok, and then if I was squaring \[\sqrt{x+9}\] it will become just x+9, right?

- phi

yes
and because we have an equation
we also square the right side to keep things equal

- horsegirl27

ok, and what's that going to be? 25, since it wasn't in a square root?

- phi

yes
\[ \left(\sqrt{x+9}\right)^2 =5^2 \\ x+9= 25 \]

- horsegirl27

Then I do -9, to get x=16

- horsegirl27

-9 on both sides

- phi

yes. so that means we have a radical equation
\[ 3\sqrt{x+9} + 7=22 \]
that does not have an extraneous solution (it has a "true" solution)
to make an equation with an extraneous solution, start over with
\[ \sqrt{25} = -5 \]
notice if we square both sides we get
\[ \left( \sqrt{25}\right)^2 = -5 \cdot -5 \\ 25 = 25 \]
because -5 , along with +5, is a square root of 25

- horsegirl27

ok

- horsegirl27

So it doesn't matter if something is negative or positive to be a square root?

- phi

a real number has two square roots, and you might see
\[ \pm \sqrt{x} \]
to remind us
but to continue
do the same thing as before
let 25 be x+9
let a=3, and multiply both sides by 3

- phi

\[ \sqrt{25} = -5 \\ \sqrt{x+9 } = -5\\3 \sqrt{x+9 } = -5\cdot 3 \]

- phi

and add 7 to both sides to get
\[3 \sqrt{x+9 } +7 = -8 \]

- horsegirl27

ok

- phi

can you solve this radical equation ?

- horsegirl27

I think so, let me work on it

- horsegirl27

So, first do -7 on both sides and get \[3\sqrt{x+9}=-15\] right? then divide by 3 the same way as before and get \[\sqrt{x+9}=-5\] and then square it to get x+9=25 again. Right?

- phi

yes, looks good

- horsegirl27

ok, so it's basically the same.

- phi

yes, you get x=16
now "put in 16" for x in the original equation. what do you get ?

- horsegirl27

the original equation being \[3\sqrt{x+9}+7=15?\]

- phi

this one
\[ 3 \sqrt{x+9 } +7 = -8 \]

- horsegirl27

ok

- phi

put in 16 for x on the left side, and simplify the left side. what do you get ?

- horsegirl27

I'm not quite sure how I should simplify it

- phi

the left side is
\[ 3 \sqrt{x+9 } +7 \]
replace x with 16:
\[ 3 \sqrt{16+9 } +7 = -8 \]
order of operations: do the stuff in side the radical first

- phi

in other words add 16+9

- horsegirl27

ok, that's 25.

- phi

yes, you now have
\[ 3 \sqrt{25 } +7 \]
now do the square root next

- horsegirl27

5

- phi

ok, so it's now
\[ 3 \cdot 5 +7 \]
order of operations: multiply before add

- horsegirl27

15+7
22

- phi

yes, and we just found
\[ 3 \sqrt{x+9 } +7 = -8 \]
with x= 16 becomes
\[ 22 = -8 \]

- phi

and that is just goofy.
in other words, x=16 is an "extraneous solution"
(it is not a correct solution)

- horsegirl27

ok

- phi

the problem happens because when we take the square root of 25
we always choose the "principle root" +5, and that leads to 22 on the left side
if we chose -5 as the square root of 25, we would get -8 = -8
But (for whatever reason, I am not sure why), when evaluating a radical expression we only choose the positive square root, unless there is a \( \pm\) sign in front of the square root.

- horsegirl27

ok

- phi

You are now done with this question

- horsegirl27

http://prntscr.com/7eygkr that's the next part, but we already finished that in checking it and you explained what extraneous means

- horsegirl27

thank you so much for helping me!!!!

- phi

ok, but be careful. They want the first equation to be the one with the extraneous solution.
we did it the other way round.
so use
\[ \text{ Eq 1 } 3 \sqrt{x+9 } +7 = -8 \\ \text{ Eq 2 } 3 \sqrt{x+9 } +7 = 22 \]

- horsegirl27

yup, I noticed that

- horsegirl27

thanks again, I finally understand radical equations!!

- phi

yw

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