# Find the largest positive real root

#### anemone

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

#### Opalg

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

#### anemone

##### MHB POTW Director
Staff member
$7x\sqrt{x+1} = 2x^2+3x + 3$. 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. 