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For example this equation (1): $$\sqrt{4x^2}-x+2=0$$ which then expanded to (2) $$3x^2+4x-4=0$$ and the roots via quadratic formula to (2) is $-2$ and $2/3$, but when put into (1), they don't agree with the first equation. What is the explanation behind this?

Jean Marie
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Anonymous Mouse
  • 127

3 Answers3

1

We have that since $4x^2 \ge 0$

$$\sqrt{4x^2}-x+2=0 \iff \sqrt{4x^2}=x-2 \iff (\sqrt{4x^2})^2=(x-2)^2$$

$$\iff 4x^2=x^2-4x+4 \iff 3x^2+4x-4=0 $$

is true only when $x-2\ge 0$.

Therefore the solution we obtain in this way are only valid for $x\ge 2$.

For $x<2$ we have that the original equality never can be true.

user
  • 154,566
1

I don't really see any issue whatsoever with using the standard resolution method. $$\sqrt{4x^2}-x+2=0\iff \sqrt{4x^2}=x-2\iff \begin{cases}x-2\ge 0\\ 4x^2=(x-2)^2\end{cases}\iff \begin{cases}x\ge 2\\ 3x^2+4x-4=0\end{cases}\\\iff \begin{cases}x\ge 2\\ x=\frac{-2-\sqrt{4+12}}{3}\end{cases}\lor \begin{cases}x\ge 2\\ x=\frac{-2+\sqrt{4+12}}{3}\end{cases}\iff\begin{cases}x\ge 2\\ x=-2\end{cases}\lor \begin{cases}x\ge 2\\ x=\frac{2}{3}\end{cases}$$

Therefore there aren't any solutions.

When you don't use the correct method, which involves transforming a single equation $\sqrt A=B$ into a mixed system of inequations which evaluate the signs of $A$ and $B$ before squaring both sides, you sometimes end up with a finite number of so-called spurious solutions, which do not solve the originale equation. Some other, more unlucky, times you might end up with infinitely many spurious solutions.

1

The basic explanation is that some equation transformations introduce extra solutions. E.g. if you turn

$$x=1$$

to

$$x^2=1,$$

you introduce the solution $x=-1$.

Similarly, some transformations can make you drop a solution. E.g. if you simplify

$$x^2=x$$ in

$$x=1,$$ you lose the solution $x=0$.