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2011-02-01T00:00:01
<mdoxml version="1.0">What is the value of the integers $a$ and $b$ where $$\sqrt{8 -
4\sqrt{3}} = \sqrt a - \sqrt b ?$$ <br></br></mdoxml>
<mdoxml version="1.0"><br></br>
<p>Conratulations to Sue Liu, Jonathan and Tom of Madras College,
St Andrew's and to Sanjay of The Perse School, Cambridge for their
solutions to this problem. Here is Sanjay's solution.</p>
<p>$$\sqrt{8 -4\sqrt{3}} = \sqrt{a} - \sqrt{b}$$ The tactic I shall
employ here will be to square both sides and solve for $a$ and $b$.
$$8 -4\sqrt{3} = a - 2\sqrt{ab} + b$$. From this it is clear that
the following equations must hold</p>
<div class="math">\begin{eqnarray} a+b &=& 8 \\ 4\sqrt{3}
&=& 2 \sqrt{ab} \end{eqnarray}</div>
These simultaneous equations can be solved for $a$ and $b$. From
equation 2. $$2 \sqrt{3} = \sqrt{ab}$$ From equation 1. $$b = 8
-a$$ Substituting equation 4 into 3 gives
<div class="math">\begin{eqnarray}\\ 2 \sqrt{3} &=&
\sqrt{a(8-a)}\\ 12 &=& a(8-a) \\ a^2 - 8a + 12 &=&
0 \\ (a-6)(a-2) &=& 0 \\ a &=& 2\ or \ a = 6
\end{eqnarray}</div>
From the original equation it is clear that $a = 6$ and $b=2$
because the left-hand side must be positive. Hence $$
\sqrt{8-4\sqrt{3}} = \sqrt{6} - \sqrt{2}$$ This can easily be
generalised, as follows. $$\sqrt{x-y\sqrt{z}} = \sqrt{a} -
\sqrt{b}$$ Upon squaring both sides this yields the equations
<div class="math">\begin{eqnarray} a + b &=& x \\ ab
&=& \frac{y^2z}{4} \end{eqnarray}</div>
From equation 5, it is clear that $$b = x - a$$ Substituting this
into equation 6 gives \begin{eqnarray} a(x-a) = \frac{y^2z}{4} \\
a^2 -ax + \frac{y^2z}{4} \end{eqnarray} Using the formula for
solving quadratics, and setting $a$ to be a larger than $b$ (as in
the original equation) gives the following solutions.
<div class="math">\begin{eqnarray} a &=& \frac{x+\sqrt{x^2
- y^2z}}{2} \\b &=& \frac{x-\sqrt{x^2 -
y^2z}}{2}\end{eqnarray}</div>
Note that if the minuses in the folrmula are changed to pluses,
equations 5 and 6 will still be the same, and so you'll still get
the same answers. <br></br></mdoxml>
<mdoxml version="1.0">Square and then equate rational and irrational parts.<br></br></mdoxml>
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Absurdity Again
What is the value of the integers a and b where sqrt(8-4sqrt3) =
sqrt a - sqrt b?
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