What is an extraneous solution to the equation x-3 equals sq rt 5-x?

x - 3 = √(5 - x); square both sides, for the left side use (a - b)2 = a2 - 2ab + b2

x2 - 6x + 9 = 5 - x; add x and subtract 5 to both sides

x2 - 5x + 4 = 0; this is factorable since 4 = (-1)(-4) and (-1) + (-4) = -5

(x - 1)(x - 4) = 0; let each factor equal to zero

x - 1 = 0; x = 1

x - 4 = 0; x = 4

Check if 1 and 4 are solutions to the original equation.

1 - 3 =? √(5 - 1)

-2 =? √4 (recall the radical symbol is looking only for the positive root)

-2 = 2 false, so that 1 does not satisfy the original equation, so it is an extraneous solution.

4 - 3 =? √(5 - 4)

1 =? √1

1 = 1 true, so that 4 is a solution to x - 3 = √(5 - x).