When solving a rational equation, why is it necessary to perform a check?
When performing operations, like "squaring both sides", sometimes roots are introduced that were not part of the original relation.
e.g.
let x = 4, and only 4
square both sides
x^2 = 16
now take √
√(x^2) = √16
± x = 4
x = ±4 , we now a "new" root
I will give you another example
solve √(3x+10) = x+2
square both sides
3x+10 = x^2 + 4x + 4
x^2 + x - 6 = 0
(x+3)(x-2) = 0
x = -3 or x = 2
check: (in original)
if x=2
LS = √(6+10) = √16 = 4
RS = 2+2 = 4 , checks
if x = -3
LS = √(-9+1) = 1
RS = -3+2 = -1 , does not work
so x = 2
I still do not understand it.
When solving a rational equation, it is necessary to perform a check because there is a possibility of introducing extraneous solutions. An extraneous solution is a solution that may satisfy the given equation but does not satisfy the original problem. This can happen due to certain algebraic manipulations that are not valid for all values of the variable.
To understand why a check is necessary, let's consider a simple example.
Suppose we have the rational equation: (x + 1)/x = 2.
To solve this equation, we can start by cross-multiplying to eliminate the fraction:
(x + 1) = 2x.
Next, we simplify the equation:
x + 1 = 2x.
Now, we can solve for x by moving terms around and combining like terms:
1 = x.
According to our calculations, it seems like x = 1 is the solution. However, we need to perform a check to verify if it is indeed a solution.
So, let's substitute x = 1 back into the original equation:
[(1 + 1)/1] = 2.
Simplifying this, we get: 2 = 2.
Since both sides of the equation are equal when x = 1, we can conclude that x = 1 is a valid solution.
However, this is not always the case. In some rational equations, a solution that satisfies the equation may not satisfy the original problem. This happens when an algebraic step involved dividing by a variable or an expression that could be zero, resulting in a possible extraneous solution.
For example, let's consider the equation: (x - 3)/(x + 3) = (x + 2)/(x + 2).
By solving this equation, we find that x = -1 is a potential solution. However, if we substitute x = -1 back into the original equation, we get:
(-1 - 3)/(-1 + 3) = (-1 + 2)/(-1 + 2).
Simplifying this, we get: -4/2 = 1/1.
This simplifies to: -2 = 1.
Since the two sides of the equation are not equal, we can conclude that x = -1 is an extraneous solution that does not satisfy the original problem.
Therefore, to avoid such extraneous solutions, it is crucial to perform a check by substituting the obtained solution(s) back into the original equation and verifying its validity.