Question 37413: what is an extraneous root? Answer by fractalier(6550) (Show Source):

The best way to understand what extraneous roots are is to go through a problem where you get one. The following is a solution to a problem that I did previously which dealt with extraneous roots: First, get the radical expression alone on one side of the equation: Now, square both sides of the equation to get rid of the radical:

Square both sides (since we are squaring the equation, we have to remember to check for extraneous roots at the end) The quadratic factors because and Using the zero product rule: => or => Check for extraneous roots: True (radical sign means the positive square root by convention) So is an extraneous root. The solution set is {6}

For square root, use " sqrt ". For example, square root of 7 write as sqrt(7). Use PARENTHESES to separate and correctly present different parts of formulas, like this sqrt(x+3) for , or sqrt(x)+3 for .

Solve the equation. Identify any extraneous solutions. x= sqrt-7x+8 A. 8 is a solution to the original equation. The value –1 is an extraneous solution. B.1 and –8 are solutions. C.1 and 8 are both extraneous solutions. D.1 is a solution to the original equation. The value –8 is an extraneous solution. Answer by MathLover1(20819) (Show ...

other ways to get extraneous solutions are: 1. the value makes the denominator in the original equation equal to 0 which is not permissible. 2. the value makes the radicand in a radical expression negative when the root is even. example: square root of -1 is not permissible. cube root of -1 is. 4th root of -1 isn't, etc.

Question 1047287: What are the (1) roots, and (2) extraneous roots for this equation: x=sqrt(12-x) English: x equals the square root of 12-x

Extraneous root -- exclude it and go with the more reasonable (and correct) . John ...

Now that x is out of the square root we can solve for it. This is a quadratic equation so we want one side to be zero. Subtracting x and adding 3 we get: Next we factor (or use the Quadratic Formula). This factors fairly easily: From the Zero Product Property: x-7 = 0 or x-4 = 0 Solving these we get: x = 7 or x = 4 Now we check. Use the ...

extraneous root because in problem negative integers So x=-6 is the first integer the second is (x+1)=-6+1=-5