Advanced Algebra
perfect squares | difference of two squares | forcing desired terms
These maths problems are widespread on the Internet, and often attributed to famous university entrance exams (unlikely).
Perfect Squares
Simplify `sqrt(sqrt9 + sqrt8)`
What is the most straight-forward way of cancelling the outer square root? Well, the inverse function of a square root is squaring. That would require turning the inner two surds into a perfect square. Is that even possible? Let's start by simplifying what we can.
`sqrt(sqrt9 + sqrt8)`
`sqrt(3 + 2sqrt2)`
At this point I got quite excited, because I recognised the coefficient of the middle term of a perfect square. Remember Pascal's Triangle?
`1`
`1 1`
`1 2 1`
`1 3 3 1`
`1 4 6 4 1`
And so on. That third line has the coefficients we want. We can split the first term to create a third term, turning it into an expanded perfect square.
`sqrt(1 + 2sqrt2 + 2)`
`sqrt((1+ sqrt2)^2)`
`1 + sqrt2`
And that's it.
Comment: I do encourage my students to understand perfect squares. Being able to expand a perfect square directly (without extra lines of working) can save significant time. Being able to recognise an expanded perfect square and factorise it can be a little like working magic.
Difference of Two Squares (and substitution)
Solve `(x + 1)(x + 3)(x + 5)(x + 7) = 9`
What a lovely string of odd numbers. We can take advantage of that pattern using a substitution.
`(x + 1)(x + 3)(x + 5)(x + 7) = 9`
Let `y = x + 4`
`(y - 3)(y - 1)(y + 1)(y + 3) = 9`
`(y - 3)(y^2 - 1)(y + 3) = 9`
`(y^2 - 1)(y^2 - 9) = 9`
That worked well. Let's do it again.
Let `z = y^2 - 5`
`(z + 4)(z - 4) = 9`
`z^2 - 16 = 9`
And now the variable is in one place only, making the equation much easier to solve.
`z^2 = 25`
`z = +-sqrt25`
`z = +-5`
`y^2 - 5 = +-5`
`y^2 = 0` or `y^2 = 10`
`y = 0` or `y = +-sqrt10`
`x + 4 = 0` or `x + 4 = +-sqrt10`
`x = -4` or `x = -4 +-sqrt10`
And that's it, three unique solutions.
Comment: For me, resisting the urge to factorise the difference of two squares when it's not actually helpful is a bigger problem than recognising that I can factorise.
Forcing Desired Terms
Solve `(x^7 + x^5 + x^3) / (x^6 + x^5 + x^4) = 3`
The high exponents on this thing look intimidating, but let's see what we can do. The first thing I notice is that every term on the left has a common factor of `x^3`. They can simply be factored out; `x` cannot be zero because the quotient would never equal `3` if it was, so we won't be losing a possible solution in the process.
`(x^7 + x^5 + x^3) / (x^6 + x^5 + x^4) = 3`
`(x^4 + x^2 + 1) / (x^3 + x^2 + x) = 3`
The numerator is one order higher than the denominator. In other words, the highest exponent of `x` on the top is one higher than the highest on the bottom. It's not far from being able to cancel out that `x^4` term. (It's generally a good to deal with the highest order term first.)
Let's force the required term in there. We can't just add an `x^3` term, but we can add and subtract a pair of `x^3` terms because doing so won't change what we have, but it will give us what we need to get rid of the `x^4` term.
`(x^4 + x^2 + 1 + x^3 - x^3) / (x^3 + x^2 + x) = 3`
`(x^4 + x^3 + x^2 + 1 - x^3) / (x^3 + x^2 + x) = 3`
`(x^4 + x^3 + x^2) / (x^3 + x^2 + x) + (1 - x^3) / (x^3 + x^2 + x) = 3`
`(x(x^3 + x^2 + x)) / (x^3 + x^2 + x) + (1 - x^3) / (x^3 + x^2 + x) = 3`
`x + (1 - x^3) / (x^3 + x^2 + x) = 3`
That's actually looking much nicer. Plough on? Force some more terms in there, this time to deal with the `-x^3`.
`x + (1 - x^3 - x^2 + x^2 - x + x) / (x^3 + x^2 + x) = 3`
`x + (- x^3 - x^2 - x + x^2 + x + 1) / (x^3 + x^2 + x) = 3`
`x + (- x^3 - x^2 - x) / (x^3 + x^2 + x) + (x^2 + x + 1) / (x^3 + x^2 + x) = 3`
`x - (x^3 + x^2 + x) / (x^3 + x^2 + x) + (x^2 + x + 1) / (x(x^2 + x + 1)) = 3`
`x - 1 + 1 / x = 3`
OK, that worked a little better than expected. Now it looks like a quadratic.
`x - 4 + 1 / x = 0`
`x^2 - 4x + 1 = 0`
From here it's straight forward to solve by completing the square.
`(x - 2)^2 - 4 + 1 = 0`
`(x - 2)^2 = 3`
`x - 2 = +-sqrt3`
`x = 2 +-sqrt3`
Comment: Forcing terms to turn a numerator into a multiple of a denominator is not a particularly common technique at highschool level, but it can be very handy for some integration and partial fraction questions.
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