Simplifying the Square Root of (a + b*sqrt(c))Date: 07/29/2004 at 12:44:01 From: Diego Subject: More advanced simplification of square roots I want to know how to simplify more advanced square roots, such as sqrt(11 - 2sqrt(18)). I know that the answer is 3 - sqrt(2), and that it is related to the fact that 9*2 = 18 and 9 + 2 = 11, but I don't understand why. Thank you very much, Diego Date: 07/29/2004 at 17:45:51 From: Doctor Vogler Subject: Re: More advanced simplification of square roots Hi Diego, Thanks for writing to Dr Math. That's a good question, and it interested me enough, when I was not much older than you, to figure out how it works. So we start with a number of the form X = sqrt( a + b * sqrt(c) ) and we suppose that all squares in c have been factored out into b, so that we would say that c is "squarefree." (For example, we would change sqrt(12) to 2*sqrt(3).) Now we want to write that as a sum of square roots of integers (or perhaps rational numbers). Sometimes it happens that you will get a new square root, as in sqrt( 12 + 4 * sqrt(5) ) = sqrt(2) + sqrt(10) So the first thing I did was to convince myself that if X could be written as a sum of square roots of rational numbers, then it would either be in the form X = r + s * sqrt(c) or in the form X = r * sqrt(d) + s * sqrt(c*d) = sqrt(d) * ( r + s * sqrt(c) ) I did this by noticing that the square of a sum of square roots like r + s * sqrt(c) + t * sqrt(d) will always end up with square root terms not found in a + b * sqrt(c) I won't go through all the boring details of this, but will instead get right to the heart of the problem. We know that if X is a sum of square roots of rational numbers, then X = sqrt(d) * ( r + s * sqrt(c) ) either for d=1 (where that term disappears) or for some other d. And we want this to equal X = sqrt( a + b * sqrt(c) ) So we square both of them and get X^2 = d*r^2 + d*s^2*c + 2*d*r*s * sqrt(c) X^2 = a + b * sqrt(c) Now, since c has no rational square root, these two things can only be equal if the coefficients of the sqrt(c)'s are equal, and the other parts are equal, so d*r^2 + d*s^2*c = a 2*d*r*s = b From these two equations, we find the following remarkable fact: a^2 - b^2*c = (d*r^2 + d*s^2*c)^2 - (2*d*r*s)^2 * c = d^2*r^4 + 2*d^2*r^2*s^2*c + d^2*s^4*c^2 - 4*d^2*r^2*s^2c = d^2*r^4 - 2*d^2*r^2*s^2*c + d^2*s^4*c^2 = (d*r^2 - d*s^2*c)^2 What this means is that if X can be written as a sum of square roots like we want, then a^2 - b^2*c has to be the square of a rational number. This doesn't usually happen, but if it doesn't, then that means you can stop looking for a square root of that form. In your case, it IS a square. Now let's suppose that we calculate a^2 - b^2*c = m^2 and find that it is the square of a rational number m. Then that means d*r^2 + d*s^2*c = a 2*d*r*s = b d*r^2 - d*s^2*c = m Actually, it really means that |d*r^2 - d*s^2*c| = m (those are absolute value bars) but we will consider the two possibilities separately. In fact, they work out almost the same, so I'll just do one of them here, the one I already mentioned. The other case results in a sign changing, and that will appear in my summary. Now add a+m, and subtract a-m: a+m = 2*d*r^2 a-m = 2*d*s^2*c So if we divide by 2 and take the square roots, then we will get sqrt((a+m)/2) = r * sqrt(d) sqrt((a-m)/2) = s * sqrt(c*d) and these are exactly the two terms we wanted! So now let me summarize our findings: Suppose we want to write sqrt( a + b * sqrt(c) ) as a sum of square roots of rational numbers. Then we calculate a^2 - b^2*c and check it it is the square of a rational number. If its square root is not rational, then you can't write your square root as a sum of square roots of rational numbers. On the other hand, if its square root is m, so that m^2 = a^2 - b^2*c then if b is positive, sqrt( a + b * sqrt(c) ) = sqrt((a+m)/2) + sqrt((a-m)/2) and if b is negative, sqrt( a + b * sqrt(c) ) = sqrt((a+m)/2) - sqrt((a-m)/2) Unfortunately, there isn't such a nice formula when you want higher roots like cube roots inside of cube roots, so I don't have any advice for those, but it will solve the problem you gave me. It also has the remarkable feature that it works when you want the square root of a sum of four terms. The following works because the steps above can be stated in more generality. Suppose I want to find the square root of 42 - 16 * sqrt(3) - 2 * sqrt(5) First I treat the sqrt(3)'s as rational, and look at [42 - 16 * sqrt(3)] - 2 * sqrt(5) To write this as a sum of two square roots, I would need [42 - 16 * sqrt(3)]^2 - 2^2 * 5 to be a square. So that means that I need the square root of [42 - 16 * sqrt(3)]^2 - 2^2 * 5 = 2512 - 1344 * sqrt(3) That doesn't look very nice, but we know how to take square roots of things like that. If that is a sum of two square roots, then we would need 2512^2 - 1344^2 * 3 = 891136 to be a square of an integer. And so we punch it in our calculator and hit the square root button. And voila! It is! 891136 = 944^2 So we calculate sqrt( 2512 - 1344 * sqrt(3) ) = sqrt((2512+944)/2) - sqrt((2512-944)/2) = 24 * sqrt(3) - 28 Amazing! We're on a roll! Now where were we? We were trying to get the square root of a^2 - b^2*c = [42 - 16 * sqrt(3)]^2 - 2^2 * 5 and we did. So next we calculate (a+m)/2 = ([42 - 16 * sqrt(3)] + [24 * sqrt(3) - 28])/2 = 7 + 4 * sqrt(3) (a-m)/2 = ([42 - 16 * sqrt(3)] - [24 * sqrt(3) - 28])/2 = 35 - 20 * sqrt(3) Now we know that our number is sqrt( 42 - 16 * sqrt(3) - 2 * sqrt(5) ) = sqrt((a+m)/2) - sqrt((a-m)/2) = sqrt( 7 + 4 * sqrt(3) ) - sqrt( 35 - 20 * sqrt(3) ) So now we just need to take the square roots of those two things. We're almost there! 7^2 - 4^2 * 3 = 1 = 1^2 so m = 1, and (a+m)/2 = (7+1)/2 = 8/2 = 4 (a-m)/2 = (7-1)/2 = 6/2 = 3 and sqrt( 7 + 4 * sqrt(3) ) = sqrt(4) + sqrt(3) = 2 + sqrt(3) Finally, 35^2 - 20^2 * 3 = 25 = 5^2 so m = 5, and (a+m)/2 = (35+5)/2 = 40/2 = 20 (a-m)/2 = (35-5)/2 = 30/2 = 15 and therefore sqrt( 35 - 20 * sqrt(3) ) = sqrt(20) - sqrt(15) = 2 * sqrt(5) - sqrt(3) * sqrt(5) and we end up with our final answer: sqrt( 42 - 16 * sqrt(3) - 2 * sqrt(5) ) = 2 + sqrt(3) - 2 * sqrt(5) + sqrt(3) * sqrt(5) Isn't that amazing? Math is just wonderful. Be warned: You have to be careful of your signs in all of this, but you can always just fall back on this policy: If it doesn't work, then try changing the sign. If you have any questions about this or need more help, please write back, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/