# Why don't the roots of an equation agree with the equation sometimes? [duplicate]

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?

• How exactly do you get from (1) to (2)? – Martin R Oct 21 '19 at 8:48
• @ Martin R. - The first equation is equivalent to $\sqrt{4x^2} = x-2$. Square both sides, and then bring everything to the left-hand-side. – Eevee Trainer Oct 21 '19 at 8:49
• @EeveeTrainer: You are right, I didn't see that. – However, this must have been asked and answered multiple times. – Martin R Oct 21 '19 at 8:52

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.

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.

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 $$x=0$$.