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Find the largest positive real root

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Find the largest positive real solution to the equation $7x\sqrt{x+1}-3=2x^2+3x$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
$7x\sqrt{x+1} = 2x^2+3x + 3$. 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 $\frac12\bigl(9+3\sqrt{13}\bigr) \approx 9.908.$
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
$7x\sqrt{x+1} = 2x^2+3x + 3$. 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 $\frac12\bigl(9+3\sqrt{13}\bigr) \approx 9.908.$
Thank you for participating, Opalg...your answer is correct and your method is neat.:)