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False solution to equation involving square roots
By Monalisa Chati (P2218) on Saturday, July 1, 2000 - 06:36 pm:

Hi There!

The following equation is a bit confusing one. So can anyone among

let me know why this is so?



Ö
 
4x-3
 
 +
Ö
 
2x+3
 
 = 6
By solving the above equation, we get the solutions x = 3 and x = 111

But, when we substitute the value of "x = 111", it does not come equal to 6.

Then why is "x = 111" not satisfying the equation even though it is a solution of the equation and also satisfies as per the sq. root conditions.

Monalisa
By Andrew Smith (P2517) on Sunday, July 2, 2000 - 12:39 pm:

Monalisa, the problem arises in that square roots can be positive and negative, i.e. sq. root (25) = 5 but also sq. root (25) = -5. So the two solutions you found could be solutions of: I solved the equation by squaring both sides, getting the square root by itself and squaring again then solving the quadratic equation. Doing this with any of the four equations above gives the same answer, basically because a negative number squared is positive.

I hope that has helped.

By Dan Goodman (Dfmg2) on Sunday, July 2, 2000 - 01:45 pm:

Andrew has pointed out the basic problem, which is that square roots can be positive or negative, I'll just add one thing. Starting from your equation:


Ö
 
4x-3
 
 +
Ö
 
2x+3
 
 = 6

You can get to:

x2-114x+333=0 ... (2)

by doing algebraic operations on equation (1), like adding, subtracting, squaring, etc.

If you have an equation, and you want to form another equation with the same solutions by doing algebraic operations to both sides of the equation, you have to make sure the operations are reversible. For instance, adding 5 to both sides is reversible, because you can subtract 5 from both sides to get back to the original equation. However, squaring both sides is not reversible, because square root can be positive or negative. For instance, starting from the equation x=-x (which has only one solution, x=0), squaring both sides gives you x2=(-x)2=x2, every number is a solution to this equation.

However, everything is not lost, because any solution to (1) must also be a solution to (2), even if the steps are not reversible! So, to find the solutions to (1), work out all the solutions to (2) and now all you need to do is check which of these are also solutions to (1). Therefore, 3 is the only solution.
By Monalisa Chati (P2218) on Wednesday, July 5, 2000 - 05:34 pm:

Thanks a lot for helping me guys!

Monalisa