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how do you solve algebraically and then check the potential soluation and why it isn't a soluation

 Problem 1)     √4 -    x     =     -2   

                              x -2           x -2

Problem 2)     √x + 2 - x = 0

Was confused on the answers; please re-advise

 

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1 Answer

sqrt(4) = 2.

For 2 - x/(x-2) = -2/(x-2)   start by multiplying through by (x-2) to get the variable out of the denominator.  This gives:

2*(x-2) - x = -2.  Expand this out, then combine the constant terms on one side and variable terms on the other.

2*x - 2*2 - x = -2

2*x - x = 4 - 2    Factor out the coefficients in front of x

(2-1)*x = 2   Divide to isolate x

x = 2

If you set x = 2 in the original equation, the result is to divide by 0 so it is not a valid solution.

 

For sqrt(x) + 2 - x = 0, I like to do this with a substitution by defining another variable y = sqrt(x).

The new equation becomes

y + 2 - y2 = 0, multiply by -1

y2 -y - 2 = 0

(y-2)(y+1) = 0

y = 2, -1

Since y = sqrt(x), x = y^2, so:

x = 4, 1

However, plugging these back into the original equation shows that only x =4 is a valid solution.  This is because the sqrt(x) cannot equal (-1).