A solution to an equation that you get at the end of whatever method you use that does not actually solve the original equation. One well-known example:
1=2 ====>0=0 Therefore, one equals two.
x0 x0
The laws of algebra says that we can do this because we multiplied both sides by zero. Logically, we all know this isn't actually true. This is what extraneous solutions look like when solving linear equations:
2x+3=9 If you assume x=1... 2(1)+3=9 ...and multiply everything by 0...
0=0. Therefore, my guess is correct and x=1. <==This is extraneous
This is a minor flaw in algebra, but not like what Russel's paradox in set theory because there are explanations why this happens. Here's a more useful example of how extraneous roots occur naturally:
Solve this the way every algebra student knows, and you get x=-2. Put this back into the equation, however, and you divide by zero. -2 is an extraneous solution in this case. If the only solutions are extraneous, then the equation cannot be solved.
This is why we check rational equations: we want only actual solutions.
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No. Sometimes they are both extraneous.
An "extraneous solution" is not a characteristic of an equation, but has to do with the methods used to solve it. Typically, if you square both sides of the equation, and solve the resulting equation, you might get additional solutions that are not part of the original equation. Just do this, and check each of the solutions, whether it satisfies the original equation. If one of them doesn't, it is an "extraneous" solution introduced by the squaring.
well there are two types...no solution or empty set(slashed 0) or in algebra two extraneous solutions...problems that seem to have answers but if you check they don't. its easier to give an example of the no solution: -2x-3=-2x+6 +2 +2 __________ -3 does not = 6 so there is no solution remember all variables have to cancel out and the integer cant equal each other for this to occur. if either everything including the integers canceled out or both sides equal the same then it is all real numbers. hope this helps :)
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).
It can happen. Then there is no solution!It can happen. Then there is no solution!It can happen. Then there is no solution!It can happen. Then there is no solution!