The largest positive real solution to the equation \(7x\sqrt{x+1}-3=2x^2+3x\) is \(\frac{1}{2}(9+3\sqrt{13}) \approx 9.908\). This solution is derived by squaring both sides of the equation, leading to the polynomial \(4x^4 - 37x^3 - 28x^2 + 18x + 9 = 0\), which factors into \((x^2 - 9x - 9)(4x^2 - x - 1) = 0\). The positive roots obtained are \(\frac{1}{2}(9+3\sqrt{13})\) and \(\frac{1}{8}(1+\sqrt{17})\), with the former being the larger root. Verification confirms that this root satisfies the original equation.
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Opalg said:[sp]$7x\sqrt{x+1} = 2x^2+3x + 3$. Square both sides: $49x^2(x+1) = \bigl(2x^2+3x + 3\bigr)^2 = 4x^4 + 12x^3 + 21x^2 + 18x + 9$, so that $4x^4 - 37x^3 - 28x^2 + 18x + 9 = 0.$ That factorises as $\bigl(x^2 - 9x - 9\bigr)\bigl(4x^2 - x - 1\bigr) = 0.$ The positive roots are $\frac12\bigl(9+3\sqrt{13}\bigr)$ and $\frac18\bigl(1+ \sqrt{17}\bigr)$. The first of those is the larger. But it came from squaring the original equation, so we have to check that it satisfies that equation and was not introduced by squaring. It does satisfy the original equation, so the answer is $\frac12\bigl(9+3\sqrt{13}\bigr) \approx 9.908.$[/sp]