What is an extraneous solution and why does it happen

When you solve an equation, you often transform it into a simpler form. If the transformation is reversible for every number, the solution sets stay identical. However, some operations, such as squaring both sides, lose information about sign, allowing numbers that never satisfied the initial equation to appear as solutions.

Why extraneous solutions appear

Squaring a both‑sides statement replaces (a = b) with ($a^2$ = $b^2$). The new equation is true for both (a = b) and (a = -b). Consequently, any value that makes the original sides opposite in sign will satisfy the squared equation but not the original. Similar loss of equivalence occurs when raising to an even power, multiplying by an expression that could be zero, or clearing a denominator without checking for zero.

Operations that can introduce extraneous roots:

• Squaring both sides of an equation
• Raising both sides to any even integer power
• Multiplying by a factor that might be zero
• Cross‑multiplying fractions without domain checks

Consider the radical equation $\sqrt{x+3}=x-1$. Square both sides: $x+3 = (x-1)^2$ which expands to $x+3 = x^2 -2x +1$. Rearranging gives $x^2 -3x -2 = 0$. Factoring yields $(x-2)(x+1)=0$, so the candidates are $x=2$ and $x=-1$. Substituting back, $\sqrt{2+3}=\sqrt{5}\approx2.24$ while $2-1=1$; not equal, so $x=2$ fails. For $x=-1$, $\sqrt{-1+3}=\sqrt{2}\approx1.41$ and $-1-1=-2$; also not equal. Both candidates are extraneous, showing that squaring introduced solutions that do not satisfy the original radical equation.

Procedure to eliminate extraneous solutions:

1. 1Solve the transformed equation as usual.
2. 2List all candidate solutions.
3. 3Plug each candidate back into the original equation.
4. 4Discard any candidate that does not satisfy the original.

Summary of the example: