Question 1048230: Consider the following equation:
[β/( β – 2)] + 2 = (2β – 2)/( β – 2)
Is β = 2 a root of the original equation? If not, why? Discuss the importance of excluding
values that make a denominator 0 when solving equations.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! [b/( b – 2)] + 2 = (2b – 2)/( b – 2)
Over a common denominator of (b-2), we have
b+2(b-2)=2b-2, cancelling out the denominator
b+2b-4=2b-2
b=2
That root always has to be substituted into the original equation.
Here, it causes a denominator to be 0 and is extraneous. There is no solution to this equation. When numbers are squared, when denominators may disappear in the algebra, in the end, the result obtained must be checked in the original equation. A negative x cannot be a value for sqrt (x) in the original equation, and b=2 is not a root here.
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