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The value of √(3-2 √2) is
Answers
Answer:
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Step-by-step explanation:
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Explanation:
The answer above is completely correct, but it may help the student to see how to reach it:
We seek a square root of
3−2√2
. This must have the form
a+b√2
.(a+b√2)
2=a2+2ab√2+2b2
So a2+2b2=3
and
2ab=−2⇒ab=−1
These are two simultaneous equations for
a and b
.Rearrange the second:
b=−1a
Substitute into the first:
a2+2a2=3
a4+2=3a2
a4−3a2+2=0
Note that this is a quadratic in
a2 :(a2)2−3(a2)+2=0
Factorise:
(a2−2)(a2−1)=0
This gives us two possible solutions for
a2 : 2 and 1 , and so the four solutions for
a : ±√2 and ±1
We are looking for integer solutions for
a , and so ±1
are possible solutions. But the other two are possible too - they can simply be folded in to the
√2
term. This wouldn't have been possible if we'd had the root of some other number in the solution for
a
, but this solution is a special case.
Now use the second equation to deduce the four equivalent solutions for
b :b=−1
ab=¯¯¯¯¯+1√2=¯¯¯¯¯+12√2
and
¯¯¯¯+1
.So we have the four solution pairs
(a,b) :(√2,−12√2)
(−√212√2)
(1,−1)(−1,1)
This is a bit suspicious - we expect only two solutions, positive and negative square roots, so we wonder if some of these are identical to each other: When we substitute them in to our desired expression
a+b√2
, we get:√2−12√2√2=−1+√2 −√2+12√2√2=1−√2
1−√2−1+√2
So the two solutions with
√2
are identical to the two simpler solutions, so we can get rid of them. We now have two solutions, positive and negative square roots:
1−√2−1+√2
When we take the written square root of a quantity, it is implied that the desired root is the positive root, the "principal value" of the square root function. So we take the single solution that comes from
a=+1 :1−√2
Double check: Make sure that this produces the desired answer:
(1−√2)2=1−2
√2+2=3−2√2
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