an extraneous solution.
1) When solving radical equations, it is often convenient to square both sides of the equation. 2) When doing this, extraneous solutions may be introduced - the new equation may have solutions that are not solutions of the original equation. Here is a simple example (without radicals): The equation x = 5 has exactly one solution (if you replace x with 5, the equation is true, for other values, it isn't). If you square both sides, you get: x2 = 25 which also has the solution x = 5. However, it also has the extraneous solution x = -5, which is not a solution to the original equation.
Details may vary depending on the equation. Quite often, you have to square both sides of the equation, to get rid of the radical sign. It may be necessary to rearrange the equation before doing this, after doing this, or both. Squaring both sides of the equation may introduce "extraneous" roots (solutions), that is, solutions that are not part of the original equation, so you have to check each solution of the second equation, to see whether it is also a solution of the first equation.
The solution set is the answers that make an equation true. So I would call it the solution.
That's the "solution" of the equation.
Extraneous solution
an extraneous solution.
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.
No. Sometimes they are both extraneous.
Extraneous means extra and unnecessary. Extraneous solutions are values that can arise from the process of solving the equation but do not in fact satisfy the initial equation. These solutions occur most often when not all parts of the process of solving are not completely reversible - for example, if both sides of the equation are squared at some point.
extraneous solution. or the lines do not intersect. There is no common point (solution) for the system of equation.
That's an extraneous solution. You need to check for these when algebraically solving equations, especially when you take both sides of an equation to a power.
1) When solving radical equations, it is often convenient to square both sides of the equation. 2) When doing this, extraneous solutions may be introduced - the new equation may have solutions that are not solutions of the original equation. Here is a simple example (without radicals): The equation x = 5 has exactly one solution (if you replace x with 5, the equation is true, for other values, it isn't). If you square both sides, you get: x2 = 25 which also has the solution x = 5. However, it also has the extraneous solution x = -5, which is not a solution to the original equation.
An unacceptable solution is an extraneous one.
Then it is not a solution of the original equation. It is quite common, when solving equations involving radicals, or even when solving equations with fractions, that "extraneous" solutions are added in the converted equation - additional solutions that are not solutions of the original equation. For example, when you multiply both sides of an equation by a factor (x-1), this is valid EXCEPT for the case that x = 1. Therefore, in this example, if x = 1 is a solution of the transformed equation, it may not be a solution to the original equation.
Details may vary depending on the equation. Quite often, you have to square both sides of the equation, to get rid of the radical sign. It may be necessary to rearrange the equation before doing this, after doing this, or both. Squaring both sides of the equation may introduce "extraneous" roots (solutions), that is, solutions that are not part of the original equation, so you have to check each solution of the second equation, to see whether it is also a solution of the first equation.
This usually refers to a word problem. For example, say you have a rectangle with an area of 6 and sides of length (x-3) and (x+2) and you have to find the value of x. Solving said equation, you find that x=-3 or 4. Substituting these values back into the (x-3) and (x+2), you find that if x=-3, one of the sides of the rectangle would equal −6. Since you can't have a distance of -6 (distance is always positive), that answer is considered extraneous. In essence, an extraneous solution is one that does not make sense in the context of the problem.