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horsegirl27
 one year ago
Please help me out, will medal!! I'm really confused. It's creating a radical equation. I do not understand radical equations at all.
horsegirl27
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.1Could someone explain radical equations, then help me make 2 for this question? And what does extraneous mean?

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1@phi @ParthKohli @mathmate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01) 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
 one year ago
Best ResponseYou've already chosen the best response.1How do I isolate it?

phi
 one year ago
Best ResponseYou've already chosen the best response.3I think I would start by picking a number and squaring it. for example 5 and 5*5= 25 and write \[ \sqrt{25} = 5 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3they 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 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.3ok. 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

phi
 one year ago
Best ResponseYou've already chosen the best response.3If it is not clear, we are creating a complicated equation that matches what they want \[ a \sqrt{x+b} + c = d \]

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1I understand how we're replacing numbers for the letters

phi
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.3almost 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
 one year ago
Best ResponseYou've already chosen the best response.1okay, should we use any number for c?

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, we are making it up. It can be anything (but 0 would be cheating)

phi
 one year ago
Best ResponseYou've already chosen the best response.3\[ 3\sqrt{x+9} + c=15+c \\ 3\sqrt{x+9} + 7=15+7 \\ 3\sqrt{x+9} + 7=22 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.1Do you know what extraneous means?

phi
 one year ago
Best ResponseYou've already chosen the best response.3Extraneous 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
 one year ago
Best ResponseYou've already chosen the best response.3Let's try to solve your equation \[ 3\sqrt{x+9} + 7=22 \] First add 7 to both sides

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1that will take out +7 and change 22 to 15

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1\[3\sqrt{x+9}=15 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3now divide both sides by 3

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1how do I do that? I know 15 will become 5, but what about the left side?

phi
 one year ago
Best ResponseYou've already chosen the best response.3I 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

phi
 one year ago
Best ResponseYou've already chosen the best response.3so you get \[ \frac{3 \sqrt{x+9}}{3} = \frac{15}{3} \]

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1Wait, after simplifying, if I divide 3 by 3, will that be changing it to 1?

phi
 one year ago
Best ResponseYou've already chosen the best response.3on the left , the 3/3 = 1 on the right the 15/3 = 5

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1ok, I knew the right side

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, 3/3 =1 and 1 times anything is the anything

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1So it will be \[1\sqrt{x+3}=5\] or is that wrong?

phi
 one year ago
Best ResponseYou've already chosen the best response.3btw, notice \[ \frac{15}{3} = \frac{3 \cdot 5}{3} = \frac{3}{3} \cdot 5 =1\cdot 5= 5 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3You 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
 one year ago
Best ResponseYou've already chosen the best response.1Ohhhh yes it finally makes sense now!

phi
 one year ago
Best ResponseYou've already chosen the best response.3another 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 \[ \left( \frac{1}{3} \cdot 3 \right) \cdot 3 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3and definitely not \[ \left( \frac{1}{3} \cdot 3 \right) \cdot\left( \frac{1}{3} \cdot 3 \right) \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3but back to the problem. what do you have so far?

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1um I think \[1\sqrt{x+9}=5\]

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, but generally we don't bother to show the multiply by 1

phi
 one year ago
Best ResponseYou've already chosen the best response.3in other words we know 1 times \(\sqrt{x+9}\) is \(\sqrt{x+9}\) so we just write \[ \sqrt{x+9}=5 \]

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1ok, that makes sense

phi
 one year ago
Best ResponseYou've already chosen the best response.3we 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
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, we don't need to write all of that

phi
 one year ago
Best ResponseYou've already chosen the best response.3to "get rid" of the radical operator, square both sides what do we get?

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1well, 9 squared is 81, and 5 squared is 25

phi
 one year ago
Best ResponseYou've already chosen the best response.3you have \[ \sqrt{x+9}=5 \] "square both sides" means do \[ \left(\sqrt{x+9}\right)^2 =5^2\]

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1ohhhhhhhhhhhhhhhhhhhhhhhh

phi
 one year ago
Best ResponseYou've already chosen the best response.3square means "multiply by itself one time" \[ \left(\sqrt{x+9}\right)\left(\sqrt{x+9}\right) =5 \cdot 5 \]

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1that makes sense now lol

phi
 one year ago
Best ResponseYou've already chosen the best response.3by 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) )

phi
 one year ago
Best ResponseYou've already chosen the best response.3or, just remember, squaring a square root undoes the square root

phi
 one year ago
Best ResponseYou've already chosen the best response.3examples: \[ \left( \sqrt{3x}\right)^2 = 3x \\ \left( \sqrt{x^2+y^2}\right)^2 = x^2 + y^2 \]

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1that sort of makes sense

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1Okay, reading it a few times, I understand

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1You square a square root so it's no longer a square root.

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1ok, and then if I was squaring \[\sqrt{x+9}\] it will become just x+9, right?

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes and because we have an equation we also square the right side to keep things equal

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1ok, and what's that going to be? 25, since it wasn't in a square root?

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes \[ \left(\sqrt{x+9}\right)^2 =5^2 \\ x+9= 25 \]

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1Then I do 9, to get x=16

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes. 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
 one year ago
Best ResponseYou've already chosen the best response.1So it doesn't matter if something is negative or positive to be a square root?

phi
 one year ago
Best ResponseYou've already chosen the best response.3a 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
 one year ago
Best ResponseYou've already chosen the best response.3\[ \sqrt{25} = 5 \\ \sqrt{x+9 } = 5\\3 \sqrt{x+9 } = 5\cdot 3 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3and add 7 to both sides to get \[3 \sqrt{x+9 } +7 = 8 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3can you solve this radical equation ?

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1I think so, let me work on it

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1So, 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?

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1ok, so it's basically the same.

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, you get x=16 now "put in 16" for x in the original equation. what do you get ?

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1the original equation being \[3\sqrt{x+9}+7=15?\]

phi
 one year ago
Best ResponseYou've already chosen the best response.3this one \[ 3 \sqrt{x+9 } +7 = 8 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3put in 16 for x on the left side, and simplify the left side. what do you get ?

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1I'm not quite sure how I should simplify it

phi
 one year ago
Best ResponseYou've already chosen the best response.3the 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
 one year ago
Best ResponseYou've already chosen the best response.3yes, you now have \[ 3 \sqrt{25 } +7 \] now do the square root next

phi
 one year ago
Best ResponseYou've already chosen the best response.3ok, so it's now \[ 3 \cdot 5 +7 \] order of operations: multiply before add

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, and we just found \[ 3 \sqrt{x+9 } +7 = 8 \] with x= 16 becomes \[ 22 = 8 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.3and that is just goofy. in other words, x=16 is an "extraneous solution" (it is not a correct solution)

phi
 one year ago
Best ResponseYou've already chosen the best response.3the 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.

phi
 one year ago
Best ResponseYou've already chosen the best response.3You are now done with this question

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1http://prntscr.com/7eygkr that's the next part, but we already finished that in checking it and you explained what extraneous means

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1thank you so much for helping me!!!!

phi
 one year ago
Best ResponseYou've already chosen the best response.3ok, 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
 one year ago
Best ResponseYou've already chosen the best response.1yup, I noticed that

horsegirl27
 one year ago
Best ResponseYou've already chosen the best response.1thanks again, I finally understand radical equations!!
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