Ab Surd Ity
1 out of 3
Find the value of sqrt(2+sqrt3)-sqrt(2-sqrt3)and then of cuberoot(2+sqrt5)+cuberoot(2-sqrt5).
Problem
The absurdity of it!
Find the exact values of $$\sqrt{2 + \sqrt 3}- \sqrt{2 - \sqrt3}$$ and of $$\root 3 \of {2 + \sqrt 5}+ \root 3 \of {2 - \sqrt 5}$$
Getting Started
For the first part try squaring the whole expression.
For the second part, you could write $a = {\root 3\of {2+\sqrt{5}}}$ and $b= {\root 3\of {2-\sqrt{5}}}$, and use the identity $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + 3ab(a + b) + b^3.$$
Student Solutions
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!
Teachers' Resources
Why do this problem?
The problem provides good practice in the manipuation of surds and in algebra (involving the expansions of $(p+q)^2$ and $(p+q)^3$, the difference of two squares and the use of the Remainder Theorem to factorise a cubic equation). If care is taken with the algebra the result comes out in a satisfyingly neat way. The question looks complicated but it turns out to be simple.
Possible approach