Simplifying and solving surds with two variables

In summary, the conversation discussed finding values for x and y in the equation 2x-y+\sqrt{4x-y}=x-2y+3+\sqrt{x+5}. The attempt at a solution involved rearranging the equation and trying to eliminate or equate the square roots on the left-hand side. It was noted that squaring both sides of an equation does not necessarily preserve the equality, so special care must be taken. Ultimately, it was found that equating both sides was the correct approach and any solutions found for (\sqrt{4x - y} - \sqrt{x+5})^2 = (-x - y +3)^2 would also be solutions for the original equation.
  • #1
miniradman
196
0

Homework Statement


Find values for x, and y in the following statement:

[itex]2x-y+\sqrt{4x-y}=x-2y+3+\sqrt{x+5}[/itex]

The Attempt at a Solution


I've managed to rearrange the equation, but I cannot eliminate or equate the square roots on the LHS:

[itex]\sqrt{4x-y}-\sqrt{x+5}=-x-y+3[/itex]
 
Physics news on Phys.org
  • #2
Square both sides. This will give you one square root instead of two. Isolated the square root on one side. Square both sides again.

Be careful, however. It is of course true that if [itex]x=y[/itex], then [itex]x^2 = y^2[/itex]. So if you square both sides of an equality, then the equality is preserved. But the converse is not true. If [itex]x^2 = y^2[/itex], then you can not deduce [itex]x=y[/itex].

So let's say you want to solve [itex]\sqrt{x}=x-2[/itex]. Then you can square both sides and get [itex]x = x^2 - 4x + 4[/itex]. If we solve this then we get [itex]x=4[/itex] and [itex]x=1[/itex].
So what we have shown is that if [itex]\sqrt{x}=x-2[/itex], then [itex]x=4[/itex] or [itex]x=1[/itex]. We do not have the converse. Solutions of [itex]x=x^2 - 4x + 4[/itex] are not necessarily solutions of [itex]\sqrt{x}=x-2[/itex].

So any solution of [itex]\sqrt{4x - y} - \sqrt{x+5} = -x - y +3[/itex] is also a solutions of [itex](\sqrt{4x - y} - \sqrt{x+5})^2 = (-x - y +3)^2[/itex]. But not conversely. So even if you manage to solve [itex](\sqrt{4x - y} - \sqrt{x+5})^2 = (-x - y +3)^2[/itex], then its solutions will not necessarily be solutions of the original equation. In fact, you will have to check which solutions are also solutions of the original equation.
 
Last edited:
  • #3
As it turned out... I could have equated both sides. I think I must have made an arithmetic errors when I tried it the first 5 times or so :p

Thank you micromass for the intuition
 

1. What are surds with two variables?

Surds with two variables are expressions that involve two variables, typically represented by the letters x and y, and one or more square roots. The variables represent unknown quantities and the square roots represent irrational numbers.

2. How do you simplify surds with two variables?

To simplify surds with two variables, you need to follow the basic rules of simplifying surds. First, check if you can simplify any square roots by finding perfect square factors. Then, combine like terms by adding or subtracting coefficients of the variables. Finally, rearrange the terms in alphabetical order.

3. Can surds with two variables be solved?

Surds with two variables can be solved if they are in the form of an equation. To solve, you can use algebraic techniques such as isolating the variable or using quadratic formula. However, if the surds are in the form of an expression, they cannot be solved as they do not have a specific numerical value.

4. What is the difference between simplifying and solving surds with two variables?

Simplifying surds with two variables refers to the process of reducing the expression to its simplest form by combining like terms and finding perfect square factors. On the other hand, solving surds with two variables involves finding the value of the variables in an equation by using algebraic techniques.

5. Are there any special cases when simplifying or solving surds with two variables?

Yes, there are two special cases when dealing with surds with two variables. The first case is when the variables have the same coefficient, in which case you can factor out the coefficient and simplify the expression. The second case is when the variables have different coefficients, in which case you need to use the distributive property and then simplify the expression.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
6
Views
530
  • Precalculus Mathematics Homework Help
Replies
10
Views
1K
  • Precalculus Mathematics Homework Help
Replies
22
Views
3K
  • Precalculus Mathematics Homework Help
Replies
4
Views
540
  • Precalculus Mathematics Homework Help
Replies
1
Views
722
  • Precalculus Mathematics Homework Help
Replies
5
Views
704
  • Precalculus Mathematics Homework Help
Replies
21
Views
1K
  • Precalculus Mathematics Homework Help
Replies
1
Views
880
  • Precalculus Mathematics Homework Help
Replies
21
Views
688
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
Back
Top