12 Solve the equation (5)/(x+2)+(3)/(x^2)+2x=2 Show clear algebraic working.

To solve the given equation<br /><br />\[<br />\frac{5}{x+2} + \frac{3}{x^2+2x} = 2<br />\]<br /><br />multiply both sides by \((x+2)(x^2+2x)\) to clear the fractions:<br /><br />\[<br />5(x^2+2x) + 3 = 2(x+2)(x^2+2x)<br />\]<br /><br />Next, expand the brackets and simplify the result:<br /><br />\[<br />5x^2 + 10x + 3 = 2x^3 + 4x^2 + 4x<br />\]<br /><br />This simplifies to:<br /><br />\[<br />2x^3 - x^2 - 6x + 3 = 0<br />\]<br /><br />Rearrange the equation, since it's a cubic equation:<br /><br />\[<br />2x^3 + x^2 - 2x - 6x + 2 + 1 - 1 -3 = 0<br />\]<br /><br />And then, group the terms:<br /><br />\[<br />2x^3 + 1x^2 - 1x - 3 = (2x^2)(x + 1/2) - (3)(x + 1/2)<br />\]<br /><br />This simplifies further into the form of:<br /><br />\[<br />(2x^2 - 3)(x + 1/2) = 0<br />\]<br /><br />Setting each part of equation to zero, we get:<br /><br />\[<br />2x^2 - 3 = 0<br />\]<br /><br />This gives us \(x = \sqrt{3/2}\), which does not solve our original equation (need to ensure the original denominators don't lead to division by zero), and <br /><br />\[<br />x + 1/2 = 0<br />\]<br /><br />This gives us \(x = -1/2\).<br /><br />Solving the two valid exponential equations now: \(x^2+2x\ =\ 0\) and \(x+2\ =\ 0\).<br />We get solutions for first \(x=0, x=-2\) (ignore \(x=-2\) since denominator term \(x+2\) generate division by zero) and for second solution \(x=-2\), not valid again for same reason. And so, <br />Conclusion: The fall through solution for valid \(x\) is \(x= -1\ (\)from \(x = -1/2\) and expression-wise its \(x= - 1/2\). Have ensured original denominator doesn't cause division by zero issue.