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Simplifying square-root expressions

Worked examples of taking expressions with square roots and taking all of the perfect squares out of the square roots. For example, 2√(7x)⋅3√(14x²) can be written as 42x√(2x).

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  • blobby green style avatar for user ROSS MABRY
    Solve this, please square root of (x+ 15) + square root of (x) = 15
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      First, you are in the wrong section of lessons. You have a radical equation, not a radical expression.
      This is the section you need to be in: https://www.khanacademy.org/math/algebra2/radical-equations-and-functions

      For a problem more like yours, I would suggest you look at the 2nd problem at this link: http://www.purplemath.com/modules/solverad3.htm

      I'll get you started on your equation: √(x+15) + √(x) = 15
      1) I would move one radical to the other side. I think it is less confusing. The link above keeps them both on the same side.
      Subtract √(x): √(x+15) = 15 - √(x)
      2) Square both sides: [√(x+15)]^2 = [15 - √(x) ]^2
      3) Simplify left side. FOIL or use extended distribution on the right side to eliminate the exponents
      x + 15 = 225 - 30√(x) + x
      4) Subtract x: 15 = 225 - 30√(x)
      5) Subtract 225: -210 = - 30√(x)
      6) Divide by -30: 7 = √(x)
      7) Square both sides again: 7^2 = √(x)^2
      8) Simplify: 49 = x
      9) Check answer back in original equation to verify that it isn't an extraneous solution.

      hope this helps.
      (36 votes)
  • hopper cool style avatar for user The Math Master
    At , you state the answer: 6xz√2xz.
    However, what happened to the rule that x or z can't be negative?
    Did we assume all variables were greater than or equal to zero in the beginning?
    I know we don't need absolute value of the variable if it is x^2, or x^4. But in this case, it was just 6xz√2xz. Both x and z were singular, and if one of them were negative(and the other positive), wouldn't this answer be incorrect without a restraint?
    (7 votes)
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  • mr pants pink style avatar for user Hidayah698
    at , how do you solve the equation radical 75yz to the second power? i watched the video but i am still a little confused.
    (4 votes)
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    • mr pink green style avatar for user David Severin
      What is to the second power, √(75yz^2)? I think this is what you mean, so 75 breaks down to 25z^2 (perfect square) and 3y (non-perfect square), √(25z^2)*√(3y) = 5z√3y.
      The other choice is √(75yz)^2 in which case the square and square root cancel to give 75yz.
      (2 votes)
  • old spice man green style avatar for user Miguel McKinnon
    Hello, I'm hoping that someone can show me the way!

    I am working through the following square root simplification problem:

    Square Root of 108a^6

    I addressed it as follows:

    108 can be prime factored into: 2*2*3*3*3

    I then broke it down as:

    Sq rt of 2^2
    Sq rt of 3^3
    Sq rt of a^6

    I then came up with the following:

    2*3*a^3
    or further simplified: 6a^3

    The Khan answer (which I of course presume is correct) came up with the following simplification:

    6a^3 sq rt of 3

    It looks like rather than combining like integers and adding exponents, Khan multiplied 2*2*3*3*3= 6^2*3

    ----
    Is there a rule or a step that I'm missing here that brings me to an incorrect answer?

    Thanks in advance for your help!
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      Your approach is ok. But, the sqrt(3^3) will not equal 3. You need to look for perfect squares which would have an even exponent. Since this exponent is odd, regroup. Split the factors into sqrt(3^2) sqrt(3) = 3 sqrt(3)
      This makes your answer:
      2*3*a^3 sqrt(3)
      Multiply factors outside to get:
      6 a^3 sqrt(3)

      Hope this helps.
      (4 votes)
  • cacteye green style avatar for user stek.kaelyn
    Why does sal not multiply the numbers(2,14,5) together?? to get square root of 140a^5... can someone please explain!
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      He could do that and he kind of did when he put them all under one radical. But, he kept them in factored form because it is easier to see the perfect squares with factored form. If you muliply to 140a^5, you then need to split it up into perfect squares.
      (3 votes)
  • blobby green style avatar for user Macosmonoz Juliet Wapma
    Expand and simplify 5square root of 12 multiply square root of 6
    (2 votes)
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  • winston baby style avatar for user Aspen
    Is it all just a WHOLE LOT of simplifying, or am I missing something?
    (4 votes)
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  • blobby green style avatar for user nep1019
    So none of the videos here show why anything beyond the 5th power for a variable has a perfect square that equals x to the power of 4 squared. So one of the questions in the foundations part lists y to the 9th. And the hint (I do not see this explained anywhere in either exponents or in simplifying square roots variables at all) shows that written out as y to the 4th squared with a remaining y but gives no explanation for WHY y to the 4th is a perfect square.
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      First, consider what happens when you square variables...
      y*y = y^2
      y^2 * y^2 = y^4
      y^3 * y^3 = y^6
      y^4 * y^4 = y^8
      Notice, all the resulting exponents are even numbers.

      Now, let's look at sqrt(y^9). The exponent is odd. So, y^9 is not a perfect square. However, just like with numbers, we can split the factors apart to find perfect squares. There are more than one way to do this.
      Sal split y^9 into y^8 * y. y^8 has an even exponent. So it is a perfect square. It can split it into 2 factor groups: y^4*y^4 * y = (y^4)^2 * y. When you do sqrt(y^8), you get the y^4.

      An alternative method is to split y^9 int pairs of y's. Each pair is a perfect square:
      y^9 = y^2 * y^2 * y^2 * y^2 * y
      You can then do the square root of each pair, which will give you y*y*y*y sqrt(y) = y^4 sqrt(y).

      Hope this helps.
      (2 votes)
  • starky sapling style avatar for user Joseph SR
    can i think of this as combining like terms? its seems so.... am i wrong?
    (3 votes)
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  • leafers ultimate style avatar for user Marvyn Greco
    why is it the numbers in all the problem I do on here never ever have a perfect square root?
    (3 votes)
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Video transcript

- [Instructor] Let's get some practice. Simplifying radical expressions that involve variables. So let's say I have two times the square root of seven x times three times the square root of 14 x squared. Pause the video and see if you can simplify it. Taking any perfect squares out multiplying and taking any perfect squares out of the radical sign. Well, let's first just multiply this thing. So, we can change the order of multiplication. This is going to be the same thing as two times three times the square root of seven x times the square root of 14 x squared. And so this is going to be equal to six times and then the product of two radicals, you can view that as the square root of the product. So six times the square root of and I'll actually I'll just leave it like this. Seven times x and then let me actually factor 14. 14 is two times seven times x squared. Actually let me extend my radical sign a little bit. And the reason why I didn't multiply it out. Obviously we could've multiplied it in our head. X times x squared is x to the third. And we could've said, seven times 14 is what 98. We could've done that. But when you're trying to factor out perfect squares, it's actually easier if it's in this factored form here. Especially because, from a variable point of view you can view this as a perfect square already. And then 14's not a perfect square, seven isn't a perfect square but seven times seven is. Seven times seven is a perfect square. That is 49 of course. So let's rewrite this a little bit to see what we can do. This is going to be six times and I could write it like this. The square root of let's put all the perfect squares first so seven times seven that is 49 that's those two. X squared 49 x squared and then I could once again separate the two radicals right over here. So whatever else is left. So I've already used the seven, the seven, the x squared I have a two x left. Times two x. Hopefully you'll appreciate that these two things are equivalent. I could've put one big radical sign over 49 x squared times two x which would've been exactly what you have there, but, if you're taking the radical of the product of things, that's the same thing as the product of the radicals. It's come straight out of our exponent properties. But what's valuable about this is we now see this is six times now we can take the the square root of 49 x squared this is going to seven x square root of 49 is seven square root of x squared is going to be x and then we multiply that times the square root of two x times the square root of two x and so now we're in the home stretch. Six times seven is 42 x times the square root of two x and the key thing to appreciate is I keep using this property that a radical of products or the square root of products is the same thing as a product of the square root. So even this step that I did here, if you wanted, you could've had an intermediary step. You could've said that the square root of 49 x squared is the same thing as square root of 49 times the square root of x squared which would get us square root of 49 is seven square root of x squared is x right over there. Let's do let's do another one of these. So let's say I have square root of two a times the square root of 14 a to the third times the square root of five a. So like always, pause this video and see if you can simplify this on your own. Multiply them and then take all the perfect squares out of the radical. So let's multiply first. So this is gonna be the same thing as the square root of two times 14 times five. So let me actually just I'm just going to two and five are prime. 14 I can factor it as two times seven so this is gonna be two times, instead of 14 I'm going to write two times seven and then times five and then we have a times a to the third times a well actually let me write that as a to the fifth. We have a to the first, times a to the third, times a to the first, and the exponents you get a to the fifth. Now, what perfect squares do we have here? Well we already see a perfect square in terms of two times two and then a to the fifth isn't a perfect square if you think of terms of the variable a but you can view that as a perfect square a to the fourth times a. So let's rearrange this a little bit. And so this is going to be equal to the square root of let me put my perfect squares out front. The square root of four, two times two times a to the fourth and then let me put my non perfect squares times I have a seven a five and an a that I haven't used yet so seven times five is 35 so it's 35 a and now just like we said before, we could let me do it we could say hey look, this is the same thing as the square root of four times the square root of a to the fourth it's using exponent properties and then times the square root of 35 a. Now principle root of four is positive two. You can view this as a positive square root and then square root of a to the fourth the principle root is going to be a squared and then we're going to have that times the square root of 35 a and we're done. Let's do one more example and this time, we're going to involve two variables which as we'll see isn't that much more complicating. So let's simplify the square root of 72 x to the third z to the third so the key is can we factor 72 is not a perfect square but is there a perfect square somewhere in there? And you immediately see that if you're trying to factor it you get 36 times two and 36 of course is a perfect square. And likewise x to the third and z to the third are not each perfect squares but they each have an x squared and z squared in them. So let me rewrite this. This is the same thing This is the same thing as I can write let me put all my perfect squares up front. So I have 36 36 I'm gonna take an x squared out x squared I'm gonna take a z squared out z squared and then we're left with is we took a 36 out so we're left with a two times two and if we took an x squared out of this we're left with just an x x to the third divided by x squared is x two x and then z to the third divided by z squared is just z and you can verify this multiply this all out. You should be getting you should be getting exactly what we have here. I do that little line on the z to differentiate so it doesn't look like my twos. 36 times two is 72 x squared times x is x to the third. Z squared times z is z to the third. But now this is pretty straighforward to factor because let me just I'll do more steps than you would probably do if you were doing it on your own but that's because the whole point here is to learn. So two x z so that's just using exponent properties. And so everything here is a perfect square. This is going to be the square root of 36 times the square root of x squared times the square root of z squared which is going to be square root of 36 is principle root of 36 is six principle root of x squared is x principle root of z squared is z and we're gonna multiply that times square root of two x z and we are done.