Extraneous solutions of radical equations (article) | Khan Academy (2024)

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  • Isabella

    8 years agoPosted 8 years ago. Direct link to Isabella's post “is 0 an extraneous soluti...”

    is 0 an extraneous solution

    (20 votes)

    • Allen Yang

      8 years agoPosted 8 years ago. Direct link to Allen Yang's post “It doesn't have to be. Fo...”

      Extraneous solutions of radical equations (article) | Khan Academy (4)

      It doesn't have to be. For example, x^2+4x=0 has solutions 0 and -4 (try them)

      (23 votes)

  • Abd Hamid Mat Sain

    8 years agoPosted 8 years ago. Direct link to Abd Hamid Mat Sain's post “square root 9 should be e...”

    square root 9 should be either 3 or -3.......why should -3 be extraneous then?

    (10 votes)

    • Kim Seidel

      8 years agoPosted 8 years ago. Direct link to Kim Seidel's post “You are correct that squa...”

      Extraneous solutions of radical equations (article) | Khan Academy (8)

      Extraneous solutions of radical equations (article) | Khan Academy (9)

      You are correct that square root (9) can be 3 or -3. But, there is a rule that we always use the principle root (the positive value) unless the square root has a minus in front of it.
      sqrt(9) is asking for the principle root = 3
      - sqrt(9) is asking for the negative root = -3.

      See this video: https://www.khanacademy.org/math/algebra-basics/basic-alg-foundations/alg-basics-roots/v/introduction-to-square-roots

      (25 votes)

  • Carla Cristina Almeida

    8 years agoPosted 8 years ago. Direct link to Carla Cristina Almeida's post “In the last situation, wh...”

    In the last situation, why did they plug -1 in the equation? was it a random number?

    (12 votes)

    • weidner.mary62

      8 years agoPosted 8 years ago. Direct link to weidner.mary62's post “This got me at first too....”

      Extraneous solutions of radical equations (article) | Khan Academy (13)

      This got me at first too. In the original question, it specifically asks you to use x=-1

      (13 votes)

  • Robert Roller

    7 years agoPosted 7 years ago. Direct link to Robert Roller's post “So take the following as ...”

    So take the following as an example.

    sqrt(4x + 41) = x+5

    Originally, x must be greater than -41/4 to be valid. When we solve this equation algebraically, we get the following two solutions: X = -8 and X = 2.

    When you plug X = -8 back into the equation, you end up with sqrt(9) = -3.
    This is extraneous because we are supposed to use the principle root.

    I have three questions.
    #1: Why do we have to use the principle root? It seems logical that the sqrt(9) should be equal to both +3 and -3.
    #2: Why is this extraneous solution still within the original domain of the function, as it is > -41/4.
    #3: What happened mathematically to produce this extraneous solution. If you graph the two equations separately, they only intersect at x = 2, therefore x = 8 is not a solution. But why is this? What happened to produce this extra answer?

    (8 votes)

    • Himalaya

      7 years agoPosted 7 years ago. Direct link to Himalaya's post “1. In your example, we us...”

      Extraneous solutions of radical equations (article) | Khan Academy (17)

      1. In your example, we use the principal square root because the original problem statement says so. It says √(4x + 41), and not -√(4x + 41). By definition, the radical notation without a minus or ± sign in front of it means “the principal root”, which is always positive. The solutions to Power Equations and the solutions to Radical Equations are different things. For example, x^2 = 4 has two solutions: x = ±2, while x = √4 has only one solution: x = 2, and x = -√4 also has one solution: x = -2. We define the square root as a function, so it must have only one output for each input. If we define the function y = √x as having two solutions, then it is no longer a function.

      2. The solution x = -8 is extraneous to the original equation √(4x + 41) = x + 5. However, it is the solution to the equation -√(4x + 41) = x + 5. The expression under the radical is same in both equations, so in terms of keeping the radicand non-negative, the value -8 is OK. If we take the function y = √(4x + 41), then -8 would be a valid input value for x. However, for the equation √(4x + 41) = x + 5, the value x = -8 is not a solution, because it leads to an invalid statement 3 = -3, which is not true.

      3. As mentioned above, x = -8 is the solution to the equation -√(4x + 41) = x + 5. So, if we graph -√(4x + 41) instead of √(4x + 41), it will intersect with x + 5 at x = -8.

      (12 votes)

  • 1earth4ever

    6 years agoPosted 6 years ago. Direct link to 1earth4ever's post “I am having a really hard...”

    I am having a really hard time with this unit. I have watched all the videos several times and I am still very confused. I would really appreciate some help.

    (7 votes)

    • Ian Pulizzotto

      6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “The key idea for solving ...”

      Extraneous solutions of radical equations (article) | Khan Academy (21)

      The key idea for solving square root equations is to isolate a radical on one side and square both sides. If there's still a radical in the equation, then this process would need to be performed a second time. (By the way, don't forget to include the middle term when squaring a binomial. Many students forget to do this.)

      After you have solved the resulting linear or quadratic equation for x, remember that you're not finished yet! Because every positive number has a positive and a negative square root, but the radical symbol denotes only the positive (principal) square root, the act of squaring both sides can create invalid (extraneous) solutions! So plug in your solutions to the original equation to determine which solutions work and which solutions must be discarded.

      Have a blessed, wonderful day!

      (10 votes)

  • Tetsuya

    7 years agoPosted 7 years ago. Direct link to Tetsuya's post “I still don't understand ...”

    I still don't understand why I should care about extraneous solution. It's outside the domain, not a solution, a wrong answer. Or is there any use of finding it?

    (6 votes)

    • kubleeka

      7 years agoPosted 7 years ago. Direct link to kubleeka's post “Extraneous solutions are ...”

      Extraneous solutions of radical equations (article) | Khan Academy (25)

      Extraneous solutions are not necessarily outside the domain. But they can appear as extra solutions when we square both sides of an equation, because when we square an equation, we would get the same result whether the original equation was positive or negative. So one solution corresponds to the positive equation, and the other to the negative equation. But both of them will fall out of the algebra. We need to be able to tell which solution is extraneous and which works.

      (10 votes)

  • Sally

    4 years agoPosted 4 years ago. Direct link to Sally's post “Hi I don't understand why...”

    Hi I don't understand why √9 would not equal -3 when it equals 3. In practice question 3, why wouldn't x=-1 be a correct solution??

    (5 votes)

    • Esther Hernandez

      4 years agoPosted 4 years ago. Direct link to Esther Hernandez's post “I think they're looking f...”

      I think they're looking for the principal sqrt of 9.

      (5 votes)

  • Joslynkim75

    5 years agoPosted 5 years ago. Direct link to Joslynkim75's post “In the video titled, "Equ...”

    In the video titled, "Equation that has a specific extraneous solution." The person who did the video squared the right side of the equation wrong. (d+2x)^2 should be d^2 + 4dx + 4x^2. This is an error.

    (4 votes)

    • Kim Seidel

      5 years agoPosted 5 years ago. Direct link to Kim Seidel's post “This is a known error. A...”

      This is a known error. A correction box does pop up on the video at about

      Extraneous solutions of radical equations (article) | Khan Academy (32) 1:05

      and tells you there is an error. However, these are only visible if you aren't watching in full screen mode. So, if you find something like this, you should always check to see if there is an error message.

      (5 votes)

  • Courtney Ellwein

    a year agoPosted a year ago. Direct link to Courtney Ellwein's post “In question 4 why can't t...”

    In question 4 why can't the square root of 9 = -3 ?

    (3 votes)

    • Kim Seidel

      a year agoPosted a year ago. Direct link to Kim Seidel's post “The problem tells you tha...”

      The problem tells you that it wants the principal root (the positive root). If the problem wanted the negative root, it would have a "-" in front.
      sqrt(9) = +3, the principal root
      -sqrt(9) = -3, the negative root

      See: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:rational-exponents-radicals/x2f8bb11595b61c86:radicals/v/introduction-to-square-roots

      (7 votes)

  • Shyawn Zahid

    8 years agoPosted 8 years ago. Direct link to Shyawn Zahid's post “For the question with Ele...”

    For the question with Elena, doesn't she have to check too because it's possible to have no solutions?

    (3 votes)

    • Denis Son

      8 years agoPosted 8 years ago. Direct link to Denis Son's post “we squared the both sides...”

      we squared the both sides by an odd number. So it's not necessary to check the equation

      (2 votes)

Extraneous solutions of radical equations (article) | Khan Academy (2024)

FAQs

What is the extraneous solution of a radical equation? ›

When you square a radical equation you sometimes get a solution to the squared equation that is not a solution to the original equation. Such an equation is called an extraneous solution. Remember to always check your solutions in the original equation to discard the extraneous solutions.

Which solution to the equation is extraneous? ›

In mathematics, an extraneous solution (or spurious solution) is one which emerges from the process of solving a problem but is not a valid solution to it.

How do you check for extraneous solutions? ›

To find whether your solutions are extraneous or not, you need to plug each of them back in to your given equation and see if they work. It's a very annoying process sometimes, but if employed properly can save you much grief on tests or quizzes.

What are extraneous roots of a quadratic equation? ›

Extraneous roots are introduced when we perform operations on both sides of an equation that are not valid for all values of the variable. To identify extraneous roots, we need to plug in each solution we obtained back into the original equation and check if it satisfies the equation.

How do you find the extraneous solution of a rational equation? ›

Identify extraneous solutions by checking if the solutions found by solving the equation make the denominator of the fraction equal to zero. If it does, then the solutions are extraneous and should be rejected.

Which solution to the equation 3 a 2 2 a 4a 4 a2 4 is extraneous? ›

Expert-Verified Answer

Extraneous solutions: From the solutions of an equation, the invalid solutions are known as extraneous solutions. For a=-2 right hand side of the given equation is not defined because the denominator become 0.

How to know if a radical equation has no solution? ›

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution. Solve: To isolate the radical, subtract 1 to both sides.

What is the difference between no solution and extraneous solution? ›

parallel lines have no solution because they never intersect. No real solution implies that there could possibly be imaginary solutions. Extraneous solutions are created be using certain methods to solve the problem. They are not, never were, or never will be solutions.

Is an extraneous solution still a solution? ›

Extraneous solutions

This means that equality is maintained. However, there might still appear solutions which do not solve the original equation.

What are the two types of equations that can have extraneous solutions? ›

Since even root functions are restricted to values greater than or equal to zero, any equation involving even roots or their corresponding fractional exponent should be checked for extraneous solutions.

What causes a solution to a rational equation to be an extraneous solution? ›

What causes a solution to a rational equation to be an extraneous solution? If a solution results in zero when substituted into the denominator of the equation, the solution is extraneous.

How to solve radical equations? ›

To solve a radical equation:
  1. Isolate the radical expression to one side of the equation.
  2. Square both sides the equation.
  3. Rearrange and solve the resulting equation.

What is an extraneous solution to a radical equation? ›

In math, an extraneous solution is a solution that emerges during the process of solving a problem but is not actually a valid solution. You can only find out whether or not a solution is extraneous by plugging the solution back into the original equation.

What is an example of a radical equation? ›

A radical equation, or a radical expression, is an expression that has a radical symbol, or a square root symbol. An example of a radical equation is y={x}^(1/2).

What is the extraneous solution to the equation sqrt 2p 1/2 sqrt p 1? ›

Expert-Verified Answer

The extraneous solution to the given equation √(2p+1) + 2√p = 1 is x = 4.

What is the solution of a radical? ›

A solution in radicals or algebraic solution is a closed-form expression, and more specifically a closed-form algebraic expression, that is the solution of a polynomial equation, and relies only on addition, subtraction, multiplication, division, raising to integer powers, and the extraction of nth roots (square roots, ...

Why is an extraneous solution? ›

The reason extraneous solutions exist is because some operations produce 'extra' answers, and sometimes, these operations are a part of the path to solving the problem. When we get these 'extra' answers, they usually don't work when we try to plug them back into the original problem.

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