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Absurdity Again

What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?


Find the smallest numbers a, b, and c such that: a^2 = 2b^3 = 3c^5 What can you say about other solutions to this problem?

Route to Root

A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?

Ab Surd Ity

Age 16 to 18 Challenge Level:

Congratulations to Hyeyoun from St Paul's Girls School, London, to Sue Liu of Madras College, St Andrew's, Scotland, Sanjay from The Perse School, Cambridge and Bill from Alcester Grammar School for your solutions.

We take the square root symbol in the question to signify the positive square root. The tactic here is to square both sides and then find the correct square root. If $\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=X$, then $$X^2= \left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right) \left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right).$$ The right hand side equals $$2 + \sqrt{3}-2\left(\sqrt{2+\sqrt{3}}\times \sqrt{2-\sqrt{3}}\right) +2-\sqrt{3},$$ and $$\sqrt{2+\sqrt{3}}\times\sqrt{2-\sqrt{3}}=1.$$ Therefore $$X^2 = 2+\sqrt{3}-2+2-\sqrt{3}=2.$$ Does $X=-\sqrt{2}$ or $+\sqrt{2}$?

Well $2+\sqrt{3}> 2-\sqrt{3}$, so $\sqrt{2+\sqrt{3}}> \sqrt{2-\sqrt{3}}$, so $X$ is positive and we have $X=\sqrt{2}$.

Note that we could take each square root to be positive or negative. If so, then the question is much harder and there are more solutions for $X$. For example, we could take $\sqrt{3} = 1{\cdot}732\cdots$; then $\sqrt{2+\sqrt{3}}$ has two values (approximately $\pm 1{\cdot}93$), and $\sqrt{2-\sqrt{3}}$ has two values (approximately $\pm 0{\cdot}52$). It follows that $X$ has four values (approximately $\pm 2{\cdot}45$ and $\pm 1{\cdot}41$). Alternatively, we could take $\sqrt{3} = -1{\cdot}732\cdots$, and then we would get even more solutions.

In the second part there are again many solutions (because square roots have two values and cube roots have three values). To simplify the solution we restrict ourselves to real cube roots. We want to find $$X = {\root 3\of {2+\sqrt{5}}} + {\root 3\of {2-\sqrt{5}}}.$$ One way to do this is to write $a = {\root 3\of {2+\sqrt{5}}}$ and $b= {\root 3\of {2-\sqrt{5}}}$, and use the equation $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + 3ab(a + b) + b^3.$$ As $X = a + b$, we have $$X^3 = (2 + \sqrt{5}) + 3X{\root 3\of {(2 + \sqrt{5})(2 - \sqrt{5})}} +(2 - \sqrt{5}).$$ As $\root 3\of {(2 + \sqrt{5})(2 - \sqrt {5}} = \root 3\of {-1} = -1$ this gives $X^3 + 3X - 4 = 0$ and hence $$(X - 1)(X^2 + X + 4) = 0.$$ As $X^2 + X + 4 = 0$ has only complex solutions, and we are looking for the real values of $X$, so we have $X = 1$, that is $${\root 3\of {2 + \sqrt{5}}} + {\root 3\of {2 - \sqrt{5}}} = 1.$$ You can check this on your calculator!