horsegirl27
  • horsegirl27
Please help me out, will medal!! I'm really confused. It's creating a radical equation. I do not understand radical equations at all.
Mathematics
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horsegirl27
  • horsegirl27
Please help me out, will medal!! I'm really confused. It's creating a radical equation. I do not understand radical equations at all.
Mathematics
jamiebookeater
  • jamiebookeater
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horsegirl27
  • horsegirl27
http://prntscr.com/7exg84
horsegirl27
  • horsegirl27
Could someone explain radical equations, then help me make 2 for this question? And what does extraneous mean?
horsegirl27
  • horsegirl27

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anonymous
  • 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
  • horsegirl27
How do I isolate it?
phi
  • 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
  • horsegirl27
Okay.
phi
  • 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
  • horsegirl27
16 and 9?
phi
  • 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
  • horsegirl27
3
phi
  • phi
If it is not clear, we are creating a complicated equation that matches what they want \[ a \sqrt{x+b} + c = d \]
horsegirl27
  • horsegirl27
I understand how we're replacing numbers for the letters
phi
  • 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
  • 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
  • horsegirl27
okay, should we use any number for c?
phi
  • phi
yes, we are making it up. It can be anything (but 0 would be cheating)
horsegirl27
  • horsegirl27
Okay let's use 7.
phi
  • phi
\[ 3\sqrt{x+9} + c=15+c \\ 3\sqrt{x+9} + 7=15+7 \\ 3\sqrt{x+9} + 7=22 \]
phi
  • 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
  • horsegirl27
Do you know what extraneous means?
phi
  • 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
  • phi
Let's try to solve your equation \[ 3\sqrt{x+9} + 7=22 \] First add -7 to both sides
horsegirl27
  • horsegirl27
that will take out +7 and change 22 to 15
horsegirl27
  • horsegirl27
\[3\sqrt{x+9}=15 \]
phi
  • phi
now divide both sides by 3
horsegirl27
  • horsegirl27
how do I do that? I know 15 will become 5, but what about the left side?
phi
  • 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
  • horsegirl27
ohhh ok
phi
  • phi
so you get \[ \frac{3 \sqrt{x+9}}{3} = \frac{15}{3} \]
phi
  • phi
which simplifies to ?
horsegirl27
  • horsegirl27
Wait, after simplifying, if I divide 3 by 3, will that be changing it to 1?
phi
  • phi
on the left , the 3/3 = 1 on the right the 15/3 = 5
horsegirl27
  • horsegirl27
ok, I knew the right side
phi
  • phi
yes, 3/3 =1 and 1 times anything is the anything
horsegirl27
  • horsegirl27
So it will be \[1\sqrt{x+3}=5\] or is that wrong?
phi
  • phi
btw, notice \[ \frac{15}{3} = \frac{3 \cdot 5}{3} = \frac{3}{3} \cdot 5 =1\cdot 5= 5 \]
phi
  • 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
  • horsegirl27
Ohhhh yes it finally makes sense now!
phi
  • 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
  • horsegirl27
ok
phi
  • phi
and definitely not \[ \left( \frac{1}{3} \cdot 3 \right) \cdot\left( \frac{1}{3} \cdot 3 \right) \]
phi
  • phi
but back to the problem. what do you have so far?
horsegirl27
  • horsegirl27
um I think \[1\sqrt{x+9}=5\]
phi
  • phi
yes, but generally we don't bother to show the multiply by 1
horsegirl27
  • horsegirl27
ok
phi
  • phi
in other words we know 1 times \(\sqrt{x+9}\) is \(\sqrt{x+9}\) so we just write \[ \sqrt{x+9}=5 \]
horsegirl27
  • horsegirl27
ok, that makes sense
phi
  • 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
  • horsegirl27
Yeah, we don't need to write all of that
phi
  • phi
to "get rid" of the radical operator, square both sides what do we get?
horsegirl27
  • horsegirl27
well, 9 squared is 81, and 5 squared is 25
phi
  • phi
you have \[ \sqrt{x+9}=5 \] "square both sides" means do \[ \left(\sqrt{x+9}\right)^2 =5^2\]
horsegirl27
  • horsegirl27
ohhhhhhhhhhhhhhhhhhhhhhhh
phi
  • phi
square means "multiply by itself one time" \[ \left(\sqrt{x+9}\right)\left(\sqrt{x+9}\right) =5 \cdot 5 \]
horsegirl27
  • horsegirl27
that makes sense now lol
phi
  • 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
  • horsegirl27
ok
phi
  • phi
or, just remember, squaring a square root undoes the square root
phi
  • phi
examples: \[ \left( \sqrt{3x}\right)^2 = 3x \\ \left( \sqrt{x^2+y^2}\right)^2 = x^2 + y^2 \]
horsegirl27
  • horsegirl27
that sort of makes sense
phi
  • phi
It is a definition
horsegirl27
  • horsegirl27
Okay, reading it a few times, I understand
horsegirl27
  • horsegirl27
You square a square root so it's no longer a square root.
phi
  • phi
yes
horsegirl27
  • horsegirl27
ok, and then if I was squaring \[\sqrt{x+9}\] it will become just x+9, right?
phi
  • phi
yes and because we have an equation we also square the right side to keep things equal
horsegirl27
  • horsegirl27
ok, and what's that going to be? 25, since it wasn't in a square root?
phi
  • phi
yes \[ \left(\sqrt{x+9}\right)^2 =5^2 \\ x+9= 25 \]
horsegirl27
  • horsegirl27
Then I do -9, to get x=16
horsegirl27
  • horsegirl27
-9 on both sides
phi
  • 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
  • horsegirl27
ok
horsegirl27
  • horsegirl27
So it doesn't matter if something is negative or positive to be a square root?
phi
  • 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
  • phi
\[ \sqrt{25} = -5 \\ \sqrt{x+9 } = -5\\3 \sqrt{x+9 } = -5\cdot 3 \]
phi
  • phi
and add 7 to both sides to get \[3 \sqrt{x+9 } +7 = -8 \]
horsegirl27
  • horsegirl27
ok
phi
  • phi
can you solve this radical equation ?
horsegirl27
  • horsegirl27
I think so, let me work on it
horsegirl27
  • 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
  • phi
yes, looks good
horsegirl27
  • horsegirl27
ok, so it's basically the same.
phi
  • phi
yes, you get x=16 now "put in 16" for x in the original equation. what do you get ?
horsegirl27
  • horsegirl27
the original equation being \[3\sqrt{x+9}+7=15?\]
phi
  • phi
this one \[ 3 \sqrt{x+9 } +7 = -8 \]
horsegirl27
  • horsegirl27
ok
phi
  • phi
put in 16 for x on the left side, and simplify the left side. what do you get ?
horsegirl27
  • horsegirl27
I'm not quite sure how I should simplify it
phi
  • 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
  • phi
in other words add 16+9
horsegirl27
  • horsegirl27
ok, that's 25.
phi
  • phi
yes, you now have \[ 3 \sqrt{25 } +7 \] now do the square root next
horsegirl27
  • horsegirl27
5
phi
  • phi
ok, so it's now \[ 3 \cdot 5 +7 \] order of operations: multiply before add
horsegirl27
  • horsegirl27
15+7 22
phi
  • phi
yes, and we just found \[ 3 \sqrt{x+9 } +7 = -8 \] with x= 16 becomes \[ 22 = -8 \]
phi
  • phi
and that is just goofy. in other words, x=16 is an "extraneous solution" (it is not a correct solution)
horsegirl27
  • horsegirl27
ok
phi
  • 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
  • horsegirl27
ok
phi
  • phi
You are now done with this question
horsegirl27
  • 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
  • horsegirl27
thank you so much for helping me!!!!
phi
  • 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
  • horsegirl27
yup, I noticed that
horsegirl27
  • horsegirl27
thanks again, I finally understand radical equations!!
phi
  • phi
yw

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