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How Do You Know If A Solution Is Extraneous

How Do You Know If A Solution Is Extraneous. Up to 10% cash back extraneous solutions. To tell if a solution is extraneous you need to go back to the original problem and check to see if it is actually a solution.

Which Solution is Extraneous?? from brainly.com

To tell if a solution is extraneous you need to go back to the original problem and check to see if it is actually a solution. But 2 is excluded from the domain of the original equation because it would make the denominator of one of the fractions. But this can't be a solution as both denominators are zero when x is 1.

X = Sqrt(2X+3) If You Square Both Sides And Solve For X You.

To tell if a solution is extraneous you need to go back to the original problem and check to see if it is actually a solution. This equation has no solution. But this can't be a solution as both denominators are zero when x is 1.

Notice The Highest Degree Of This Polynomial Is 1.

Solve for x , 1 x − 2 + 1 x + 2 = 4 ( x − 2) ( x + 2). We've got the study and writing resources you need for your assignments. If you don't, that solution is extraneous.

This Equation Has No Solution.

X = x + 2. If the solution is extraneous it won’t make sense. Up to 10% cash back extraneous solutions.

An Extraneous Solution Is A Root Of A Transformed Equation That Is Not A Root Of The Original Equation Because It Was Excluded From The Domain Of The Original Equation.

Simply plug the solutions you get back into your original equation. To tell if a solution is extraneous you need to go back to the original problem and check to see if it is actually a solution. Solution for how do you tell if a solution is an extraneous solution?

They Are Extraneous Because They Are Not Solutions Of The Original Problem.

But 2 is excluded from the domain of the original equation because it would make the denominator of one of the fractions. To illustrate my point, let’s pick a problem with an extraneous solution. To tell if a solution is extraneous you need to go back to the original problem and check to see if it is actually a solution.

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