# 2007 iTest Problems/Problem 55

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The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

## Problem

Let $R=675$. Let $x$ be the smallest real solution of $3x^2+Rx+R=90x\sqrt{x+1}$. Find the value of $\lfloor x\rfloor$.

## Solution

Plugging in $R$ results in \begin{align*} 3x^2 + 675x + 675 &= 90x\sqrt{x+1} \\ x^2 + 225x + 225 &= 30x\sqrt{x+1} \end{align*} We can solve the equation a bit easier by letting $a = x+1$. \begin{align*} x^2 + 225a &= 30x\sqrt{a} \\ x^4 + 450x^2 a + 225^2 a^2 &= 900x^2 a \\ (x^2 - 225a)^2 &= 0 \\ x^2 - 225x - 225 &= 0 \end{align*} Using the Quadratic Formula yields $x = \tfrac{225 \pm 15\sqrt{229}}{2}$. However, we have to check for extraneous solutions and calculate the highest integer less than or equal to $x$. This can be easily done using a calculator, but in this solution, we will not use a calculator.

We know that $15.1^2 = 228.01$ and $15.15^2 = 229.5225$, so $226.5 < 15\sqrt{229} < 227.25$. That means $-1.5 < \tfrac{225 - 15\sqrt{229}}{2} < -1.125$. Inserting the value into the equation would result in the right hand side having a square root of a negative number, so $\tfrac{225 - 15\sqrt{229}}{2}$ is an extraneous solution.

On the other hand, since we know that there is a solution to the equation and $\tfrac{225 + 15\sqrt{229}}{2}$ makes both sides positive in the original equation, $\tfrac{225 + 15\sqrt{229}}{2}$ is a valid solution. We know that $15\sqrt{229}$ is greater than $225$, but we need to know if it’s less than $227.$ \begin{align*} 15\sqrt{229} &\bigcirc 227 \\ 225 \cdot 229 &\bigcirc 227^2 \\ 51525 &\bigcirc 51529 \\ \end{align*} This means that $15\sqrt{229}$ is less than $227,$ so $\tfrac{225 + 15\sqrt{229}}{2}$ is less than $226$. Thus, the value of the greatest integer less than or equal to $x$ is $\boxed{225}$.