It would be much nicer to have a simple problem instead of a hard one. So a natural way to solve a hard mathematical problem is to make it simpler.

**The key principle is to simplify.**

At summer schools we often ask the simple question: how will you solve the equation \((x-2)^2=1\)? Most students expand the brackets and then apply the quadratic formula.

The standard quadratic formula is a correct general recipe. However, a simple computer program will easily beat you. While you are solving one quadratic equation, the chip in your mobile phone can solve a billion of similar equations. So mathematics is not a cookbook of recipes, but a beautiful subject where efficiency, not brute force, is highly valued.

**Efficiency often wins over brute force.**

If we replace \((x-2)^2=1\) by a very similar equation \((x^2-2x)^2=1\), then expanding the brackets will produce a longer and more complicated equation of degree 4. Moreover, any computer program based on the standard quadratic formula will fail.

However, both equations \((x-2)^2=1\) and \((x^2-2x)^2=1\) have the same pattern: a square equals a number. Hence the equations can be solved in the same way. For instance, we could replace the expression under the square by a new variable.

**A substitution is a common way to simplify.**

In the equation \((x-2)^2=1\) we could set \(y=x-2\). Then we get the simpler equation \(y^2=1\), which has two roots \(y=\pm 1\). Hence the original equation \((x-2)^2=1\) has two roots \(x=y+2=-1+2=1\) and \(x=y+2=1+2=3\).

Similarly, in the equation \((x^2-2x)^2=1\) we could set \(y=x^2-2x\). Then we get the simpler equation \(y^2=1\), \(y=\pm 1\). Hence the original equation \((x^2-2x)^2=1\) is equivalent to the union of 2 simpler quadratic equations \(x^2-2x=-1\) and \(x^2-2x=1\).

**How to simplify: try to keep a pattern.**

Most students would probably solve the last two equations by using the standard quadratic formula again. We could actually continue the simplification and complete a simple square instead. Namely, \(x^2-2x=-1\), \(x^2-2x+1=0\), \((x-1)^2=0\), so \(x=1\) is a double root. Similarly, \(x^2-2x=1\), \(x^2-2x+1=2\), \((x-1)^2=2\), \(x-1=\pm\sqrt{2}\), \(x=1\pm\sqrt{2}\).

There are many more complicated equations with the same pattern \(y^2=1\). For instance, try to solve the equations \((x^3-2x-1)^2=4\), \((\sin x+\cos x)^2=1\), \((e^{2x}-2e^x)^2=1\).

**Riddle 3**: simplify \(\sqrt{x^2}\) for any real number x and explain your conclusion.**How to submit**: to write your full answer, simply submit a comment.**Hint**: the square root \(\sqrt{x}\) of any positive real number \(x\) is always positive.**Warning**: many 2nd year maths students at a top university get it wrong.**Restriction**: only the first fully correct public answer will be rewarded.**Prize**: free 1-year access to one of our interactive web tutorials.**Update**: R S has solved the riddle, read our confirmation.

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