# How to solve hard problems: a key principle with examples 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.

If you wish to receive automatic e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.