ANSWER
A solution of the radical equation that does not satisfy the original radical equation.
EXPLANATION
An extraneous is the solution that does not satisfy the original equation.
For instance, given the radical equation:
[tex] \sqrt{x + 3} = x - 3[/tex]
We square both sides to get:
[tex] { (\sqrt{x + 3} )}^{2} = {(x - 3)}^{2} [/tex]
We expand to get;
[tex]{x + 3} = {x}^{2} - 6x + 9[/tex]
We write in standard quadratic forms:
[tex] {x}^{2} - 6x - x + 9 - 3= 0[/tex]
[tex] {x}^{2} - 7x+6= 0[/tex]
[tex] {(x - 6)(x - 1)} = 0[/tex]
This implies that;
[tex]x = 1 \: or \: x = 6[/tex]
When we substitute x= 6 into the equation, we get;
[tex] \sqrt{6+ 3} = 6- 3[/tex]
[tex] \sqrt{9} = 3[/tex]
This statement is true.
However when we substitute x=1, we get:
[tex] \sqrt{1+ 3} = 1- 3[/tex]
[tex] \sqrt{4} = - 2[/tex]This statement is false.
Hence x=1 u s referred to as an extraneous solution.
