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10) \( \frac{2 a}{x+3 a}-\frac{x}{a-x}=\frac{4 a^{2}}{x^{2}+2 a x-3 a^{2}} \)

Ask by Campbell Murphy. in South Africa
Mar 08,2025

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The solution is \( x = -6a \).

Solution

We start with the equation \[ \frac{2a}{x+3a} - \frac{x}{a-x} = \frac{4a^2}{x^2+2ax-3a^2}. \] Notice that the quadratic in the denominator on the right factors as \[ x^2+2ax-3a^2=(x+3a)(x-a). \] Thus, the equation becomes \[ \frac{2a}{x+3a} - \frac{x}{a-x} = \frac{4a^2}{(x+3a)(x-a)}. \] Since \[ a-x=-(x-a), \] we combine the two fractions on the left using the common denominator \((x+3a)(a-x)\). Write \[ \frac{2a}{x+3a} - \frac{x}{a-x} = \frac{2a(a-x) - x(x+3a)}{(x+3a)(a-x)}. \] Expand the numerator: \[ 2a(a-x)=2a^2-2ax,\quad x(x+3a)=x^2+3ax. \] Then, \[ 2a(a-x)-x(x+3a)=2a^2-2ax-x^2-3ax=2a^2-5ax-x^2. \] So the left side becomes \[ \frac{-x^2-5ax+2a^2}{(x+3a)(a-x)}. \] Recognize that \[ (x+3a)(a-x)=-(x+3a)(x-a). \] Thus, we write \[ \frac{-x^2-5ax+2a^2}{-(x+3a)(x-a)}=\frac{x^2+5ax-2a^2}{(x+3a)(x-a)}. \] Now the equation is \[ \frac{x^2+5ax-2a^2}{(x+3a)(x-a)}=\frac{4a^2}{(x+3a)(x-a)}. \] Provided that \((x+3a)(x-a)\neq 0\), we may equate the numerators: \[ x^2+5ax-2a^2=4a^2. \] Subtract \(4a^2\) from both sides: \[ x^2+5ax-6a^2=0. \] This is a quadratic in \(x\). Its discriminant is \[ \Delta=(5a)^2-4(1)(-6a^2)=25a^2+24a^2=49a^2. \] Thus, \[ x=\frac{-5a\pm\sqrt{49a^2}}{2}=\frac{-5a\pm7a}{2}. \] This gives the solutions: \[ x=\frac{-5a+7a}{2}=a\quad \text{or}\quad x=\frac{-5a-7a}{2}=-6a. \] Next, we check for extraneous solutions by considering the restrictions: \[ x+3a\neq0\quad \Longrightarrow\quad x\neq -3a, \] \[ a-x\neq0\quad \Longrightarrow\quad x\neq a. \] The solution \(x=a\) violates the condition \(a-x\neq 0\) and must be rejected. Thus, the only valid solution is \[ \boxed{x=-6a}. \]

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To solve the equation \[ \frac{2 a}{x+3 a}-\frac{x}{a-x}=\frac{4 a^{2}}{x^{2}+2 a x-3 a^{2}}, \] we start by simplifying the right-hand side. The denominator \( x^{2}+2 a x-3 a^{2} \) can be factored. Let's look for factors: \[ x^2 + 2ax - 3a^2 = (x + 3a)(x - a). \] Now, rewriting the right-hand side gives: \[ \frac{4a^2}{(x + 3a)(x - a)}. \] Now we rewrite the left-hand side to have a common denominator. The common denominator for the two fractions on the left is \((x + 3a)(a - x)\). We have: \[ \frac{2a(a - x) - x(x + 3a)}{(x + 3a)(a - x)} = \frac{2a^2 - 2ax - (x^2 + 3ax)}{(x + 3a)(a - x)}. \] So we simplify the numerator: \[ 2a^2 - 2ax - x^2 - 3ax = 2a^2 - x^2 - 5ax. \] Thus, we rephrase the equation to: \[ \frac{2a^2 - x^2 - 5ax}{(x + 3a)(a - x)} = \frac{4a^2}{(x + 3a)(x - a)}. \] Now equate the numerators as the denominators are the same (valid as long as we don't reach the undefined points due to zero in the denominator): \[ 2a^2 - x^2 - 5ax = 4a^2 \cdot \frac{(a - x)}{(x - a)}. \] Multiplying through to eliminate the fraction, we can then rearrange and solve a quadratic equation for \(x\). To summarize the solution process, one might end up with a quadratic \( Ax^2 + Bx + C = 0 \) which can be solved using the quadratic formula. If further steps in detail or numerical solutions are needed, feel free to ask!

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