Why do we get extraneous solutions? At which point during the solution process causes this?

Imagine this: There's an equation x = 2. Right now, x is only equal to two. Square both sides, and x^2 = 4. For some reason, if you want to take the square root of both sides, and you get x= +/- 2, because -2 squared is still equal to four. But, according to the original equation, x is only equal to 2. Therefore -2 is an extraneous solution, and squaring both sides of the equation creates them.

In the last problem how does it equal 30x and not 15x to begin with in step 2?

Hey there friend! So in Step 2 you get ((5+3x)^2 which is equivalent to (5+3x)*(5+3x). After you distribute you get this string of numbers: 25+15x+15x+9x^2 From there you simplify to: 9x^2+30x+25. I hope this helps! Have a good day! :D

how do u work out the the square root of n equal square root of 2n plus 6

sqrt(n) = sqrt(2n+6) To undo a square root, you raise it to the 2nd power (square it). What you do on one side you must do on the other side. So sqrt(n)^2 = n while sqrt(2n+6)^2 = 2n+6 This will leave you with n=2n+6 to solve for n.

*Is there a specific way to solve for square roots or is it always just going to be trial and error?*

We can use a linear approximation to find a close estimate for the square root. √(x) ≈ (x + y) / (2 * √(y)) where y is a number that is "close to" x. Typically, you would choose y to be a perfect square to make the math easy.

Interesting, the videos didn't discuss (that I saw) there being NO solutions.

That is covered in a later video in this lesson.

(2x + 5)(x + 5) = 0 How did you get this?

Factor the quadratic using grouping. See the videos at this link: https://www.khanacademy.org/math/algebra/polynomial-factorization#factoring-quadratics-2

For practice question #1, why couldn't I square both sides- the squareroot of 2x-7 -3 and then the 4? Wouldn't that keep the equations equivalent?

You could, but it doesn't eliminate the radical. You may be thinking that if you square the left side that it becomes 2x-7+9. It doesn't. (a+b)^2 never equals a^2+b^2. Here's what happens if you square both sides immediately [sqrt(2x-7)-3]^2 = 4^2 The left side is a binomial squared, so use FOIL or extended distribution to square that side. 2x-7 -3sqrt(2x-7) - 3sqrt(2x-7) + 9 = 16 2x+2-6sqrt(2x-7) = 16 You have just made the equation more complicated, and it still has a radical. This is why the correct 1st step is to isolate the radical *before* squaring both sides. Hope this helps.

On the left sidebar this worksheet is after the exercise "Solve Square Root Equations (basic)", but the subtitle is "Practice Some Problems Before Going Into the Exercise." Is there a mix up here? I'm accessing the page via URL: https://www.khanacademy.org/math/algebra2/radical-equations-and-functions/solving-square-root-equations/a/solving-square-root-equations-advanced

Hmm... I think it might be a typo.

For root(7x^2 + 15x) - 2x = 5 + x Got solutions as x = -5 and x = -(5/2) While checking for extraneous Solutions - For x = -5 we get root(100) + 10 = 0 The solution mentioned on the website considers only positive value of root(100) i.e. +10. Shouldn't it also consider -10 ?

There is a difference between taking the square root of a number which is always positive (√100=10) and solving x^2=100 which gives both a positive and negative answer. The first is finding a value on the square root function, the second is finding the x intercepts of an equation. Does this make sense? I looked on DESMOS and they did not cross as a check. This means both answers are extraneous.

Main content

Course: Get ready for AP® Calculus > Unit 5

Lesson 5: Square-root equations

Solving square-root equations

Google Classroom

Practice some problems before going into the exercise.

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Vincent T
Posted 5 years ago. Direct link to Vincent T's post “Why do we get extraneous ...”
Why do we get extraneous solutions? At which point during the solution process causes this?
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(12 votes)
Answer
- 🔥Cool Dude🔥
  Posted 5 years ago. Direct link to 🔥Cool Dude🔥's post “Imagine this: There's an ...”
  Imagine this: There's an equation x = 2.
  Right now, x is only equal to two.
  Square both sides, and x^2 = 4.
  For some reason, if you want to take the square root of both sides, and you get x= +/- 2, because -2 squared is still equal to four.
  But, according to the original equation, x is only equal to 2.
  Therefore -2 is an extraneous solution, and squaring both sides of the equation creates them.
  Comment on 🔥Cool Dude🔥's post “Imagine this: There's an ...”
  (54 votes)
MalichiWatson
Posted 8 years ago. Direct link to MalichiWatson's post “In the last problem how d...”
In the last problem how does it equal 30x and not 15x to begin with in step 2?
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(7 votes)
Answer
- Uri
  Posted 8 years ago. Direct link to Uri's post “Hey there friend! So in ...”
  Hey there friend!
  
  So in Step 2 you get ((5+3x)^2 which is equivalent to (5+3x)*(5+3x).
  After you distribute you get this string of numbers: 25+15x+15x+9x^2
  From there you simplify to: 9x^2+30x+25.
  
  I hope this helps! Have a good day! :D
  Comment on Uri's post “Hey there friend! So in ...”
  (9 votes)
Shadeen Higgins
Posted 8 years ago. Direct link to Shadeen Higgins's post “how do u work out the the...”
how do u work out the the square root of n equal square root of 2n plus 6
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(6 votes)
Answer
- Vu
  Posted 8 years ago. Direct link to Vu's post “sqrt(n) = sqrt(2n+6) To u...”
  sqrt(n) = sqrt(2n+6)
  To undo a square root, you raise it to the 2nd power (square it). What you do on one side you must do on the other side.
  So sqrt(n)^2 = n while sqrt(2n+6)^2 = 2n+6
  This will leave you with n=2n+6 to solve for n.
  Button navigates to signup page
  (8 votes)
s o n i a ✞
Posted 4 years ago. Direct link to s o n i a ✞'s post “*Is there a specific way ...”
*Is there a specific way to solve for square roots or is it always just going to be trial and error?*
Button navigates to signup pageComment on s o n i a ✞'s post “*Is there a specific way ...”
(6 votes)
Answer
- KLaudano
  Posted 4 years ago. Direct link to KLaudano's post “We can use a linear appro...”
  We can use a linear approximation to find a close estimate for the square root. √(x) ≈ (x + y) / (2 * √(y)) where y is a number that is "close to" x. Typically, you would choose y to be a perfect square to make the math easy.
  Button navigates to signup page
  (7 votes)
caldwelljt
Posted a year ago. Direct link to caldwelljt's post “Interesting, the videos d...”
Interesting, the videos didn't discuss (that I saw) there being NO solutions.
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(6 votes)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “That is covered in a late...”
  That is covered in a later video in this lesson.
  Button navigates to signup page
  (6 votes)
Fahmida Khanam
Posted 5 years ago. Direct link to Fahmida Khanam's post “(2x + 5)(x + 5) = 0 How d...”
(2x + 5)(x + 5) = 0 How did you get this?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “Factor the quadratic usin...”
  Factor the quadratic using grouping. See the videos at this link: https://www.khanacademy.org/math/algebra/polynomial-factorization#factoring-quadratics-2
  Button navigates to signup page
  (7 votes)
White, Kaden
Posted 4 years ago. Direct link to White, Kaden's post “For practice question #1,...”
For practice question #1, why couldn't I square both sides- the squareroot of 2x-7 -3 and then the 4? Wouldn't that keep the equations equivalent?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “You could, but it doesn't...”
  You could, but it doesn't eliminate the radical.
  
  You may be thinking that if you square the left side that it becomes 2x-7+9. It doesn't. (a+b)^2 never equals a^2+b^2.
  
  Here's what happens if you square both sides immediately
  [sqrt(2x-7)-3]^2 = 4^2
  The left side is a binomial squared, so use FOIL or extended distribution to square that side.
  2x-7 -3sqrt(2x-7) - 3sqrt(2x-7) + 9 = 16
  2x+2-6sqrt(2x-7) = 16
  You have just made the equation more complicated, and it still has a radical.
  
  This is why the correct 1st step is to isolate the radical before squaring both sides.
  
  Hope this helps.
  Comment on Kim Seidel's post “You could, but it doesn't...”
  (6 votes)
sakshi190704
Posted 3 years ago. Direct link to sakshi190704's post “For root(7x^2 + 15x) - 2x...”
For root(7x^2 + 15x) - 2x = 5 + x
Got solutions as x = -5 and x = -(5/2)
While checking for extraneous Solutions -
For x = -5 we get
root(100) + 10 = 0
The solution mentioned on the website considers only positive value of root(100) i.e. +10. Shouldn't it also consider -10 ?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- David Severin
  Posted 3 years ago. Direct link to David Severin's post “There is a difference bet...”
  There is a difference between taking the square root of a number which is always positive (√100=10) and solving x^2=100 which gives both a positive and negative answer. The first is finding a value on the square root function, the second is finding the x intercepts of an equation. Does this make sense?
  I looked on DESMOS and they did not cross as a check. This means both answers are extraneous.
  Button navigates to signup page
  (6 votes)
svmejia
Posted 8 years ago. Direct link to svmejia's post “On the left sidebar this ...”
On the left sidebar this worksheet is after the exercise "Solve Square Root Equations (basic)", but the subtitle is "Practice Some Problems Before Going Into the Exercise." Is there a mix up here? I'm accessing the page via URL: https://www.khanacademy.org/math/algebra2/radical-equations-and-functions/solving-square-root-equations/a/solving-square-root-equations-advanced
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(5 votes)
Answer
- Jimmy
  Posted 5 months ago. Direct link to Jimmy's post “Hmm... I think it might b...”
  Hmm... I think it might be a typo.
  Button navigates to signup page
  (4 votes)
jb268536
Posted a year ago. Direct link to jb268536's post “I don't know how to answe...”
I don't know how to answer these questions.
Button navigates to signup pageComment on jb268536's post “I don't know how to answe...”
(6 votes)
Answer

Course: Get ready for AP® Calculus > Unit 5

Solving square-root equations

Introduction

Practice question 1: Isolating the radical term

Step 1: Isolate the radical term.

Step 2: Take the square of both sides.

Step 3: Check for extraneous solutions.

Practice question 2: Two possible solutions

Step 1: Isolate the radical term.

Step 2: Take the square of both sides.

Step 3: Check for extraneous solutions.

Conclusion

Practice question 3

Step 1: Isolate the radical term.

Step 2: Take the square of both sides.

Step 3: Check for extraneous solutions.

Conclusion

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