Q:

What is an extraneous solution to a radical equation

Accepted Solution

A:
ANSWERA solution of the radical equation that does not satisfy the original radical equation.EXPLANATIONAn 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.