SOLUTION: pls.. help me in this problem. Use the properties of proportionality to solve: {{{a+x+sqrt(a^2-x^2)/a+x-sqrt(a^2-x^2)=b/x

Algebra ->  Proportions  -> Lessons -> SOLUTION: pls.. help me in this problem. Use the properties of proportionality to solve: {{{a+x+sqrt(a^2-x^2)/a+x-sqrt(a^2-x^2)=b/x      Log On


   



Question 550989: pls.. help me in this problem.
Use the properties of proportionality to solve:
pls..+help+me+in+this+problem.%0D%0AUse+the+properties+of+proportionality+to+solve%3A%0D%0Aa+x+sqrt(a^2-x^2)/a+x-sqrt(a^2-x^2)=b/x

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I assume your equation is
%28a%2Bx%2Bsqrt%28a%5E2-x%5E2%29%29%2F%28a%2Bx-sqrt%28a%5E2-x%5E2%29%29=b%2Fx
Multiplying both sides of the equal sign by the product of the denominators we get an (almost) equivalent equation stating that the cross products are equal. I assume that's what was meant by "Use the properties of proportionality."
x%28a%2Bx%2Bsqrt%28a%5E2-x%5E2%29%29=%28a%2Bx-sqrt%28a%5E2-x%5E2%29%29b
(Multiplying by an expression that includes variables risks introducing extraneous solutions, solutions of the new equations that are not solutions of the original equation).
x%28a%2Bx%29%2Bx%2Asqrt%28a%5E2-x%5E2%29=%28a%2Bx%29b-b%2Asqrt%28a%5E2-x%5E2%29
x%2Asqrt%28a%5E2-x%5E2%29%2Bb%2Asqrt%28a%5E2-x%5E2%29=%28a%2Bx%29b-x%28a%2Bx%29
%28x%2Bb%29%2Asqrt%28a%5E2-x%5E2%29=%28a%2Bx%29%28b-x%29
Now we square both sides of the equation. (By doing that we may be introducing extraneous solutions too).
%28x%2Bb%29%5E2%2A%28a%5E2-x%5E2%29=%28a%2Bx%29%5E2%28b-x%29%5E2








2bx%28a%5E2-x%5E2%2Ba%5E2%2Bx%5E2%29=%28x%5E2%2Bb%5E2%29%282x%5E2%2B2ax%29-4abx%5E2
2bx%282a%5E2%29=%28x%5E2%2Bb%5E2%29%282x%5E2%2B2ax%29-4abx%5E2
4a%5E2bx=2x%28x%5E2%2Bb%5E2%29%28x%2Ba%29-4abx%5E2
4a%5E2bx%2B4abx%5E2=2x%28x%5E2%2Bb%5E2%29%28x%2Ba%29
4abx%28a%2Bx%29=2x%28x%5E2%2Bb%5E2%29%28x%2Ba%29
2ab%28a%2Bx%29=%28x%5E2%2Bb%5E2%29%28x%2Ba%29
2ab=x%5E2%2Bb%5E2 --> x%5E2=2ab-b%5E2=b%282a-b%29
The expression above implies a%5E2-x%5E2=a%5E2-2ab%2Bb%5E2=%28a-b%29%5E2%3E=0, ensuring that the square roots in the original equation will have a real value.
It also results in the expression x=0+%2B-+sqrt%282ab-b%5E2%29, which includes all the solutions of the original equation, plus extraneous solutions.
If those tentative solutions were constant numbers, or simpler expressions, we could substitute and verify. In this case, it's easier to hunt down and eliminate the extraneous solutions the hard way.
Extraneous solutions include those that make the denominators zero.
There are no solutions for b=0 or b=2a, which make x zero.
For a=b, the expression found for x%5E2 turns into
x%5E2=2ab-b%5E2=a%5E2=b%5E2, and a%5E2-x%5E2=0, but x=-a makes a denominator in the original equation zero, so only x=a=b is a solution.
For sqrt%282ab-b%5E2%29-sqrt%28b%282a-b%29%29 to have a real, positive value, it must be 0%3Cb%3C2a, or 2a%3Cb%3C0.
However, there are solutions only if 0%3Ca%3Cb%3C2a, or a%3Cb%3C0.
For 0%3Cb%3Ca x%5E2=2ab-b%5E2=b%282a-b%29 gives extraneous solutions, that make just one of the factors in %28x%2Bb%29%2Asqrt%28a%5E2-x%5E2%29=%28a%2Bx%29%28b-x%29 negative.
For 2a%3Cb%3Ca%3C0, it gives extraneous solutions that make negative exactly three of the factors in %28x%2Bb%29%2Asqrt%28a%5E2-x%5E2%29=%28a%2Bx%29%28b-x%29.
Those extraneous solutions are solutions of -%28x%2Bb%29%2Asqrt%28a%5E2-x%5E2%29=%28a%2Bx%29%28b-x%29 that were introduced on squaring both sides.
THE SOLUTION
x=a=b%7D%7D%2C+when+%7B%7B%7Ba=b and
x=0+%2B-+sqrt%282ab-b%5E2%29 for 0%3Ca%3Cb%3C2a, or a%3Cb%3C0
(examples: (a,b)=(5,8), (a,b)=(-5,-2)) }.
For other (a,b) pairs there are no solutions (examples: (a,b)=(-5,-8), (a,b)=(-5,-2) ).