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Question

Solve the following for $$x$$:
$$\dfrac {1}{2a+b+2x}=\dfrac {1}{2a}+\dfrac {1}{b}+\dfrac {1}{2x}$$


Solution

$$\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\frac{1}{2x}$$

$$\dfrac{1}{2a+b+2x}+\dfrac{1}{2x}=\dfrac{2a+b}{2ab}$$

$$\dfrac{2x+2a+b+2x}{2x.(2a+b+2x)}=\dfrac{2a+b}{2ab}$$

$$\dfrac{2a+b+4x}{2ax+bx+2x^{2}}=\dfrac{2a+b}{ab}$$

$$2a^{2}b+ab^{2}+4abx=4a^{2}+2abx+4ax^{2}+2abx+b^{2}x+2bx^{2}$$

$$2a^{2}b+ab^{2}=x^{2}(4a+2b)+x(4a^{2}+b^{2})$$

$$x=\pm\left [\dfrac{-(4a^{2}+b^{2})+\sqrt{(4a^{2}+b^{2})^{2}}+4(4a+2b)(2a^{2}b+ab^{2})}{2(4a+2b)} \right]$$

Mathematics

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