Video: Pack 2 • Paper 1 • Question 21

Pack 2 • Paper 1 • Question 21

02:48

Video Transcript

Show that seven divided by three minus root two can be rewritten in the form 𝑎 plus 𝑏 root two, where 𝑎 and 𝑏 are integers.

In order to simplify or rewrite this expression, we need to multiply the top and the bottom by the conjugate of the bottom. The conjugate of three minus root two is three plus root two. The two terms remain the same. But the sign changes, either from positive to negative or in this case from negative to positive. The conjugate of 𝑎 minus root 𝑏 is 𝑎 plus root 𝑏.

Our next step is to multiply the two numerators and then separately multiply the two denominators. In order to multiply the numerator, we multiply seven by three and seven by root two. Seven multiplied by three is 21. And seven multiplied by root two is seven root two.

In order to expand the two terms on the bottom, we’ll use the FOIL method. Multiplying the first terms gives us nine. Multiplying the outside terms gives us three root two. Multiplying the inside terms gives us negative three root two. And finally, multiplying the last terms gives us negative two. This is because root two multiplied by root two is equal to root four. And the square root of four is equal to two.

As root two multiplied by root two is equal to two, then negative root two times root two is equal to negative two. We can then cancel the middle terms, as three root two minus three root two is zero.

Finally, nine minus two is equal to seven. Therefore, our dominator is seven. Seven divided by three minus root two can be rewritten as 21 plus seven root two divided by seven.

Our final step is to cancel by dividing each of the three terms by seven. This gives us a final answer of three plus root two. We’ve rewritten seven divided by three minus root two in the form 𝑎 plus 𝑏 root two, where 𝑎 equals three and 𝑏 equals one.

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