Fractional equation with a parameter

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Consider the fractional equation in the unknown $x$ depending on the parameter $a$: $\frac{1}{2x+a}-\frac{1}{a-2x}=\frac{1}{4}$

• (a) solve it for $a=3$;
• (b) find the value of $a$ for which $x=4$ is a solution.

Solution

(a) With $a=3$ the equation becomes $\dfrac{1}{2x+3}-\dfrac{1}{3-2x}=\dfrac{1}{4}$, with existence conditions $x\neq -\tfrac32$ and $x\neq \tfrac32$. The common denominator on the left is $(2x+3)(3-2x)=9-4x^2$, and the numerator is $(3-2x)-(2x+3)=-4x$: $\frac{-4x}{9-4x^2}=\frac14 \implies -16x = 9-4x^2 \implies 4x^2-16x-9=0.$ $x=\frac{16\pm\sqrt{256+144}}{8}=\frac{16\pm 20}{8}\implies x=\frac92 \ \vee\ x=-\frac12.$ Both satisfy the existence conditions. $\boxed{x=\frac92 \ \vee\ x=-\frac12}$

(b) Imposing $x=4$: $\dfrac{1}{8+a}-\dfrac{1}{a-8}=\dfrac14$. The common denominator is $a^2-64$ and the numerator $(a-8)-(8+a)=-16$: $\frac{-16}{a^2-64}=\frac14 \implies -64=a^2-64 \implies a^2=0 \implies a=0.$ Check: with $a=0$ the equation is $\dfrac{1}{2x}+\dfrac{1}{2x}=\dfrac{1}{x}=\dfrac14$, i.e. $x=4$. $\boxed{a=0}$