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Intro to solving square-root equations

Practice some problems before going into the exercise.

Introduction

In this article, we will practice the ways we can solve equations with square-roots in them.

Practice question 1

Solve the following equation for x.
2x5=7
x=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Practice question 2

Solve the following equation for x.
103x=2x+5
x=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Practice question 3

Solve the following equation for x.
5x4=x2
x=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Want to join the conversation?

  • blobby green style avatar for user jirehsm
    wow! i still dont get it
    (122 votes)
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  • male robot hal style avatar for user Manuel Huertas Luna
    When I solve a square-root equation, what is it exactly what I'm finding? In what situations would I use the number that I found? (excuse my broken english)
    (28 votes)
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  • mr pink green style avatar for user Jayson Matchado
    So, do all quadratic equations have extraneous solutions? Help pleaseeee. Does it also mean that we should always check the solutions whenever we solve for quadratics?
    (11 votes)
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    • aqualine ultimate style avatar for user Alec Traaseth
      Extraneous solutions are often introduced whenever you square both sides of an equation. After all, you're turning something that was NOT a quadratic into a quadratic. If you start out with a quadratic, you will not get extraneous solutions.

      Another situation where extraneous solutions are possible is when you're dealing with rational expressions. After cancelling denominators and simplifying down to a quadratic, you'll sometimes introduce solutions which result in a 0 in a denominator when they're plugged back in. All you have to check for in this situation is division by 0.
      (18 votes)
  • duskpin sapling style avatar for user maria.delaguardia
    i do not get it but its ok
    (7 votes)
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    • blobby purple style avatar for user Lotup
      It's been 2 months since you wrote this, so you might've gotten an answer by now, but I'll answer just in case :).

      As I understand it, when a square root is presented, it is asking for the positive square root only, unless a - or +/- sign is written to the left of the square root. However, when we solve these square-root equations, we have no way of making sure we get the solution for x that satisfies only the positive square root.

      You could also say these extraneous solutions do not make sense when you substitute them in. For example, try setting the domain of the equations before solving them.

      Let's take the example of practice question 3:

      sqrt(5x - 4) = x - 2

      Here we are saying the positive/principal square root of (5x - 4) is equal to (x - 2). That means (x - 2) has to be a positive number, which means the domain of our equation will be that x has to be greater than or equal to 2 (to ensure we get a positive number).

      One of the solutions you get when you solve practice question 3 is that x = 1. However, if x is 1 then (x - 2) will give a negative number which does not make sense for the solution and falls out of our domain (that x has to be greater than or equal to 2).

      Sorry for the long answer, and if you still don't get it, that's okay, I find it quite hard to explain maths in a block of text, so it's probably my lacking explanation in that case.

      Best of luck with maths and everything else.
      (21 votes)
  • starky tree style avatar for user hunter.curtis
    My question is "How can you do the problem without making one mistake in this long process of answer-finding?"
    (8 votes)
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  • leaf grey style avatar for user 3rdCircle
    Why does squaring work on both sides of an equation? You're multiplying each side by a different amount!
    Please clear this up for me.
    (6 votes)
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    • stelly blue style avatar for user Kim Seidel
      No, you aren't multiplying a different amount.
      Remember, the equations says that the both sides are equal to start with.
      If you square both sides, the sides are still equal.
      For example, start with: sqrt(9) = 3
      These sides are currently equal, just in a different form. Now, square both sides.
      [sqrt(9)]^2= 3^2
      You will get 9 = 9. They are still equal.
      Hope this helps.
      (14 votes)
  • blobby green style avatar for user Annaleasia DeLeon
    what if you get a desimal do you do the abc formula
    (1 vote)
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  • blobby green style avatar for user its.emma.holbrook
    i dont really understand extraneous answers... for example, in one of the problems from the video, he said 2.25 was extraneous because the square root of 2.25 = 1.5, not -1.5? but -1.5 squared = 2.25?
    (1 vote)
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    • stelly blue style avatar for user Kim Seidel
      You have forgotten that square roots have a principal root and a negative root. The default is to always use the principal root (which is the positive root). If a problem wants you to use the negative root, there will be a "-" in front of the radical.
      sqrt(2.25) = 1.5 because it is asking for the principal root
      -sqrt(2.25) = -1.5 because it is asking for the negative root

      Hope this helps.
      (11 votes)
  • blobby green style avatar for user Arnav Chauhan
    Why does
    sqrt(x-2)^2=sqrt(1) equal x-2 = +/-1 and not +/-(x-2) = +/-1 ?
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      Because some of the options create the same solution

      +(x-2) = +1 is identical to -(x-2) = -1
      -(x-2) = +1 is identical to +(x-2) = -1
      For the equation versions that match, multiply one by -1 and you get the other.
      Thus, there is no need to do 4 different sign combinations.
      Hope this makes sense.
      (4 votes)
  • aqualine seed style avatar for user ranoosh
    I did not get anything! who knows where I can find radical equations for beginners
    (4 votes)
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