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Find the rationalizing factor of a mixed surd

  1. May 6, 2014 #1
    1. The problem statement, all variables and given/known data


    $$\sqrt{2}+\sqrt{7}-\sqrt{10}$$ is a surd
    Find another irrational number such that when multiplied by $$\sqrt{2}+\sqrt{7}-\sqrt{10}$$ , it results a rationall number.

    2. Relevant equations



    3. The attempt at a solution

    I found the answer, but I think I did not find in the correct method.


    My first attempt:

    I tried multiplying $$\sqrt{2}+\sqrt{7}-\sqrt{10} with \sqrt{2}+\sqrt{7}-\sqrt{10}.$$

    I got
    [itex]
    \\
    \left( \sqrt{2}+\sqrt{7}-\sqrt{10} \right )\left (\sqrt{2}+\sqrt{7}-\sqrt{10} \right )\\
    =2+\sqrt{14}-\sqrt{20}+\sqrt{14}+7-\sqrt{70}-\sqrt{20}-\sqrt{70}+10\\
    =19+2\sqrt{14}-2\sqrt{20}-2\sqrt{70}
    [/itex]
    I didn't get a rational number
    My second attempt:

    So I tried,


    [itex]
    \\
    \left (\sqrt{2}+\sqrt{7}-\sqrt{10} \right )\left ( \sqrt{2}+\sqrt{7}+\sqrt{10} \right )\\
    =2+\sqrt{14}+\sqrt{20}+\sqrt{14}+7+\sqrt{70}-\sqrt{20}-\sqrt{70}-10\\
    =2\sqrt{14}-1\\
    [/itex]
    Extension of my second attempt

    I thought I could multiply further.

    [itex]
    \\
    \left (2\sqrt{14}-1 \right )\left (2\sqrt{14}+1\right )\\
    =4*14-1\\
    =56-1=55\\
    [/itex]
    I got a rational number by multiplying [itex]\left ( \sqrt{2}+\sqrt{7}+\sqrt{10} \right )[/itex] with

    [itex]\left (\sqrt{2}+\sqrt{7}-\sqrt{10} \right )[/itex] and further multiplying with [itex]\left (2\sqrt{14}+1\right )[/itex]

    That is, multiplying factor=
    [itex]\left ( \sqrt{2}+\sqrt{7}+\sqrt{10} \right )\left (2\sqrt{14}+1\right )[/itex]

    I think I got the answer with some guess work in my second attempt. In its extension I used
    $$a^2-b^2$$ identity.

    Is there any identity or a logical way to solve without guessing?
     
  2. jcsd
  3. May 6, 2014 #2

    Curious3141

    User Avatar
    Homework Helper

    Nothing illogical about your second approach. Using ##(a+b)(a-b) = a^2 - b^2## (multiplying by the conjugate surd) is they key to solving many of these questions. The situation is complicated here by the fact that ##a^2## is irrational, but if you remember the square of the binomial ##(x+y)^2 = x^2 + y^2 + 2xy##, you should be able to immediately see that you will reduce the number of surds from 3 to 1 with the first multiplication. All that's left is to multiply by the conjugate surd of the new expression.

    The key is always try to reduce the number of surds in each step.
     
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