Question

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

Answer 1

http://prntscr.com/7exg84

Answer 2

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

Answer 3

@phi @ParthKohli @mathmate

Answer 4

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

Answer 5

How do I isolate it?

Answer 6

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

Answer 7

Okay.

Answer 8

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 ?

Answer 9

16 and 9?

Answer 10

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

Answer 11

3

Answer 12

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

Answer 13

I understand how we're replacing numbers for the letters

Answer 14

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

Answer 15

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

Answer 16

okay, should we use any number for c?

Answer 17

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

Answer 18

Okay let's use 7.

Answer 19

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

Answer 20

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

Answer 21

Do you know what extraneous means?

Answer 22

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)

Answer 23

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

Answer 24

that will take out +7 and change 22 to 15

Answer 25

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

Answer 26

now divide both sides by 3

Answer 27

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

Answer 28

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

Answer 29

ohhh ok

Answer 30

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

Answer 31

which simplifies to ?

Answer 32

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

Answer 33

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

Answer 34

ok, I knew the right side

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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?

Answer 39

Ohhhh yes it finally makes sense now!

Answer 40

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
\[ \left( \frac{1}{3} \cdot 3 \right) \cdot 3 \]

Answer 41

ok

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

ok

Answer 47

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

Answer 48

ok, that makes sense

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

ohhhhhhhhhhhhhhhhhhhhhhhh

Answer 55

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

Answer 56

that makes sense now lol

Answer 57

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) )

Answer 58

ok

Answer 59

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

Answer 60

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

Answer 61

that sort of makes sense

Answer 62

It is a definition

Answer 63

Okay, reading it a few times, I understand

Answer 64

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

Answer 65

yes

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

Then I do -9, to get x=16

Answer 71

-9 on both sides

Answer 72

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

Answer 73

ok

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

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

Answer 78

ok

Answer 79

can you solve this radical equation ?

Answer 80

I think so, let me work on it

Answer 81

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?

Answer 82

yes, looks good

Answer 83

ok, so it's basically the same.

Answer 84

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

Answer 85

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

Answer 86

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

Answer 87

ok

Answer 88

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

Answer 89

I'm not quite sure how I should simplify it

Answer 90

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

Answer 91

in other words add 16+9

Answer 92

ok, that's 25.

Answer 93

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

Answer 94

5

Answer 95

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

Answer 96

15+7
22

Answer 97

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