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Radical and rational equations — Harder example

Watch Sal work through a harder Radical and rational equations problem.

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  • blobby green style avatar for user Neha
    lol you could've just stopped at m^2-12m+35 and did -b/a for the sum of roots, I solved this in 1 min
    (18 votes)
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  • blobby green style avatar for user wasir.alias.2014
    shouldn't the answer be 5+7=12
    (12 votes)
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  • aqualine seedling style avatar for user Aayushi :P
    At first sight of such question, I would never think about doing all of that work!
    (8 votes)
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  • female robot grace style avatar for user StudyBuddy
    Whoa, whoa, wait! @ he said that the square root of 16 is 4. But couldn't it also be negative 4? Same for the sqrt of 4 @. Couldn't that also be negative 2? How did he know to pick the positive square root?
    (5 votes)
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  • blobby green style avatar for user malakwaleed632
    why didnt we just square both sides in the beginning? why did we have to move the 3 to the other side first?
    (3 votes)
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    • piceratops ultimate style avatar for user Hecretary Bird
      The reason that Sal didn't square whole left side was because his goal in squaring was to get rid of the square root, which gets the equation into a form that we know how to work with. If we try and square the left side, when we expand it we still get a square root in our equation, which is a problem. This is because when you expand (a+b)^2, you get a perfect square trinomial a^2 + 2ab + b^2. The "2ab" term in this case would be 2(3)(sqrt(6m-26)), which in this case is ugly and has a square root in it, which makes things hard.
      Generally, you want to eliminate square roots by isolating them on one side of the equation and then squaring both sides.
      (6 votes)
  • blobby green style avatar for user Hazar Hamad Bhat
    couldn't have just used s=-b/a when turned into quadratic instead of making it a bit too long?
    (4 votes)
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  • blobby green style avatar for user jpdad06
    How should I know if to start directly adding a square root first, or if I should add the coefficients to both sides first?
    (4 votes)
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  • duskpin seed style avatar for user Tasmiya Fathima
    okay I have a doubt....at Sal writes "-6m" but according to the formula of:
    [ (a-b)^2 = a^2 - 2(a)(b) + b^2]
    so, when u put (m-3)^2 it should be equal to m^2 - 2(m)(-3) +(-3)^2 which then becomes:
    m^2 - (-6m) + 9 , which results in:
    m^2 + 6m + 9
    and due to this silly mistake, the entire solution has changed... so can someone please explain this?
    (2 votes)
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    • piceratops ultimate style avatar for user Hecretary Bird
      One small thing. When we talk about expanding out a binomial squared, you have two ways of thinking about it. Either you say:
      (a-b)^2 = a^2 - 2ab + b^2
      where b is the positive version of the number
      Or:
      (a+b)^2 = a^2 + 2ab + b^2
      where a and b can be any sign
      I like the second one better because with the first one, you can be tempted to double-count the negative. You should get something like this (first option):
      a^2 - 2ab + b^2 = m^2 - 2*m*3 + 3^2
      = m^2 - 6m + 9
      Or this (second option):
      a^2 + 2ab + b^2 = m^2 + 2*m*-3 + (-3)^2
      = m^2 - 6m + 9
      Both thought processes are perfectly fine, but I'd recommend the second one because it's easier not to double count the negative like you were doing.
      (5 votes)
  • starky ultimate style avatar for user Blade
    We could just use -b/a
    (3 votes)
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  • piceratops ultimate style avatar for user Brandon R-S
    Help! I don't understand this AT ALL! Can someone help me? Thank you!
    (1 vote)
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    • piceratops ultimate style avatar for user Hecretary Bird
      So we're asked to solve for a variable in a radical equation. Those get a bit wonky and weird, but we can square the entire equation to get out of the square root, which turns the equation into a quadratic, which we know how to solve already. Sal first isolates the square root sign so he doesn't have to square (3 - sqrt(6m - 26)) and instead can do the easier (m - 3)^2. Then, he squares both the left and right sides. From there, he puts everything on one side of the equation so we can apply the strategies for factoring and solving quadratics, getting m^2 - 12m + 35 = 0.
      From there we factor what we have. We want to have two numbers that we add to x in our factors that add to -12 and multiply to 35, so that when we multiply the factors we arrive at our original equation. These two numbers are -5 and -7. We can then write the equation as (x - 5) (x - 7) = 0. By setting each factor equal to the right side we get x = 5 and 7.
      The question asks for the sum of these solutions, which is 12. Did this help?
      (5 votes)

Video transcript

- [Instructor] What is the sum of all the solutions to the above equation? All right, this is interesting. We have three plus the square root of six m, six m minus 26 is equal to m. So anytime you have a radical equation like this, it's a good idea to try to isolate the radical. So or at least you try to isolate the radical first. So let's try to subtract three from both sides to just be left with just the radical on the left-hand side. And so three minus three is zero, and on the left-hand side, you're just left with the square root of six m minus 26, and that's going to be equal to m minus three. Now to get rid of the radical, we can square both sides of this equation. So we'll square that side, and then we can square that side, as well. And so the square root of six m minus 26 squared, that's going to be six m minus 26, and then m minus three squared, that's going to be m squared minus six m, it's gonna be minus three and minus three m plus nine. And if this step you found a little confusing, I encourage you on Khan Academy to review multiplying expressions, or I'll do a quick primer right here. M minus three times m minus three, that's the same thing as m minus three squared. You'll have m times m, which is m squared, m times negative three, which is negative three m, negative three times m, which is negative three m, and then negative three times negative three, which is positive nine. And the negative three m plus the negative three m, that is negative six m. And I encourage you, especially on like is on the SAT, you would have to do this under time pressure to be able to do things like square a binomial like this very quickly to realize it was gonna be m squared plus two times the product of these. The product of these is negative three m, two times that is negative six m, and then negative three times negative three is positive nine. So now let's see if we can solve for m. So let's get all of, let's get all of our, everything on the right-hand side of the equation. So let's subtract six m from both sides. So I'm gonna subtract six m from both sides, and let's add 26 to both sides. And the whole reason why I did that is to just clean out what I have on the left-hand side. So on the left-hand side, I'm just gonna be left with zero. And on the right-hand side, I'm gonna have m squared minus 12m plus 35, or I could write m squared minus 12m plus 35 is equal to zero. And so let's think about the m's that satisfy this. Well, I can factor this if I think about, well, what two numbers have a product of 35, but if I were to sum them, I get to negative 12? Well 35, I can think of seven times five, but seven plus five is positive 12. But what about negative seven times negative five? Well, their product is gonna be positive 35, and negative seven plus negative five is negative 12. So I could factor this into m minus five times m minus seven is equal to zero. And if this step does not make sense to you, where I factored this quadratic, encourage you to look up factoring quadratics on Khan Academy to get a little bit more practice doing that. But if I have the product of two things that equal zero, that means that one or both of them need to be equal to zero. So m minus five could be zero, or m minus seven would be equal to zero. Well, to make m minus five equal to zero, you just add five to both sides. You would have m is equal to five, or same thing over here, add seven to both sides. M is equal to seven. So the two, the two solutions are five and seven, and if you want the sum of all the solutions, it's going to be five plus seven is equal to 12. And if you just wanna verify that these actually work, try 'em out right over here. Six times five is going to be 30 minus 26, which is four. The principal root of four is positive two. Three plus two is equal to five. It's equal to m. And if m was seven, if m was seven, this is gonna be three plus the square root of 42 minus 26 is 16, and this needs to be equal to seven. Well, the principal root of 16 is four, and three plus four is indeed equal to seven.