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In this post we discuss common mistakes with square roots of real numbers. As usual in mathematics, we start from the definition: the square root \(\sqrt{x}\) is well-defined only for a non-negative real number \(x\geq 0\) and equals a unique number \(y\) such that \(x=y^2\).

The graph of the square root function \(y=\sqrt{x}\) is shown in the picture above. The function is always increasing and has the vertical tangent at the origin. If you understand derivatives, you may check that \(y'(x)\to+\infty\) as \(0<x\to 0\). We discuss only real numbers here. The square root of a complex number is another (interesting and more advanced) story.

### Easy mistake: simplifying \(\sqrt{x^2}\)

Rather surprisingly, it is a very popular myth that \(\sqrt{x^2}=x\). At least, this is the first answer we got at many summer schools and also from 2nd year maths undergraduates at a top UK university. Our next question was to check if the identity holds for \(x=-1\). Then most students realised that \(\sqrt{x^2}=x\) fails for \(x=-1\).

Some students were still in doubt and claimed that \(\sqrt{1}\) is \(1\) or \(-1\). We knew that all students have calculators at hand and asked them to compute \(\sqrt{1}\) with their calculators. The expected correct answer \(\sqrt{1}=1\) left no chance for the myth that \(\sqrt{x^2}=x\) can hold for all real \(x\).

Indeed, \(\sqrt{x^2}=|x|=\left\{\begin{array}{c} x \mbox{ for } x\geq 0,\\ -x \mbox{ for } x<0. \end{array} \right.\) At one of the summer schools we were told that the absolute value \(|x|\) is learned in the UK only at A-levels (at the age of 16-18). So there is little chance to get the correct identity \(\sqrt{x^2}=|x|\). Later is better than never, so just in case we have sketched the graph \(y=|x|\) on the left.

### Common mistake: squaring both sides

In the riddle from the post about success at Oxbridge interviews we asked to find all real \(x\) such that \(x\geq\sqrt{3x-2}\). Unfortunately, public attempts 1 and 2 (and more private attempts by e-mail) started from squaring both sides: \(x^2\geq 3x-2\).

The resulting inequality is not equivalent to the original one, because \(x=-1\) satisfies \(x^2\geq 3x-2\), but not \(x\geq\sqrt{3x-2}\). So the popular myth that \(\sqrt{x^2}=x\) has led to serious consequences that squaring both sides is a safe operation, but it is not!

When a given equation or inequality contains functions that are not well-defined for all real numbers, experts start from writing the domain when the problem makes sense. Best students always remember that a good style to finish a solution is to substitute answers back into an original equation.

However, this final check should not be considered as a part of a solution, but only as an opportunity to develop self-criticism, a key mathematical skill. Indeed, if we have infinitely many answers (as in the case of an inequality), we can not substitute back all our answers or such a substitution could be too hard. So if a final check fails, our solution fails, but the final check is ok.

For the inequality \(x\geq\sqrt{3x-2}\), we first write that square root makes sense only when \(3x-2\geq 0,\; x\geq\frac{2}{3}\). In this domain squaring both sides produces the equivalent inequality \(x^2\geq 3x-2,\; x^2-3x+2\geq 0,\; (x-1)(x-2)\geq 0\).

The last inequality has the solutions \(x\leq 1\) and \(x\geq 2\). If we remember the domain \(x\geq\frac{2}{3}\), the final answer is \(\frac{2}{3}\leq x\leq 1\) and \(x\geq 2\). Just in case, we may check that the boundary values \(\frac{2}{3},1,2\) satisfy the original inequality \(x\geq\sqrt{3x-2}\).

### Potential confusion: Q6(ii) in STEP I exam 2007

Here is the exact quote: “Given that \(x^3-y^3=(x-y)^4\) and that \(x-y=d\neq 0\), show that \(3xy=d^3-d^2\). Hence show that \(2x=d\pm d\sqrt{\frac{4d-1}{3}}\).”

Even without trying to solve the problem, any professional mathematician could spot within a few seconds that the expression \(\frac{4d-1}{3}\) under the square root can be negative. Indeed, the only given restriction \(d\neq 0\) allows \(d=-2\) when we get the complex roots \(x=-1\pm i\sqrt{3}\).

It is a good exercise in complex numbers to check that the pairs \((x,y)=(-1+i\sqrt{3},1+i\sqrt{3})\) and \((x,y)=(-1-i\sqrt{3},1-i\sqrt{3})\) actually satisfy the given equation \(x^3-y^3=(x-y)^4\), so there is no mistake in the problem. However, complex numbers are not in the STEP I syllabus and we suspect that the case of complex roots was probably missed in the original problem.

The STEP examiners’ solution contains the identity \(\pm\sqrt{9d^2+12(d^3-d^2)}=\pm d\sqrt{12d-3}\), which is luckily correct for all real d, but only if we use complex numbers with the sign \(\pm\) in both sides. For real numbers, as was probably expected, we should write \(\sqrt{|9d^2+12(d^3-d^2)|}=|d|\sqrt{|12d-3|}\), simply because the left hand side is not negative and so should the expressions under the square roots.

Unfortunately, there are other STEP problems when the sign \(\pm\) doesn’t help. So one of the homework problem from our STEP I course has a parametric equation that can’t be solved by careless squaring both sides. The average mark for this homework is less than 16/20.

Just in case the picture above shows the curve \(x^3-y^3=(x-y)^4\) excluding the line \(y=x\) of easy solutions. The straight line \(x-y=d=\frac{1}{4}\) meets the curve at the red point where \(x=\frac{1}{8}\), hence \(y=-\frac{1}{8}\). If \(d<\frac{1}{4}\), the straight line \(x-y=d\) doesn’t intersect the curve (in the real domain) and has only complex intersection points \((x,y)\), which are invisible on the usual plane.

**Riddle 12**: find all real \(x,y\) when the inequality \(\frac{x+y}{2}\geq\sqrt{xy}\) holds.**How to submit**: to write your full answer, submit a comment.**Hint**: substitute back some of your answers and non-answers.**Warning**: justify that you found all (not only some) real solutions.**Prize**: free 1-year access to one of our interactive web tutorials.**Restriction**: only the first correct public answer will be rewarded.

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This is the famous AM-GM inequality, which holds iff x,y are non-negative (given they are real).

We first show the ‘if’. Using the condition that x,y are non-negative set a^2 = x and b^2 = y. Then the inequality is (a^2+b^2)/2 >= sqrt(a^2*b^2) which simplifies to (a-b)^2 >= 0, this is clearly true.

For the ‘only if’ notice if exactly one of x,y is strictly negative the RHS would be the sqrt of a strictly negative quantity, while if both are strictly negative the LHS would be strictly negative and supposedly greater or equal to the RHS, which is strictly positive. This is clearly impossible.

Dear Carlo, your solution is almost complete. You should say something about a,b, because you have implicitly used sqrt(a^2*b^2)=ab, which doesn’t work for a=-sqrt(x), b=sqrt(y) when x,y>0. In the ‘only if’ part there are missing cases when x=0 and y<0 (or vice versa). It might be easier to first get restrictions on x,y and then consider fewer cases. You could also avoid extra variables a,b and express everything via sqrt(x), sqrt(y) for x,y>=0. We hope that you will complete your solution soon. Good luck!