Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 iTest Problems/Problem 55"

(Solution to Problem 55 — typing this on mobile took a long time)
m (→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 10: Line 10:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
3x^2 + 675x + 675 &= 90x\sqrt{x+1} \\
 
3x^2 + 675x + 675 &= 90x\sqrt{x+1} \\
−
x^2 + 225x + 225 &= 90x\sqrt{x+1}
+
x^2 + 225x + 225 &= 30x\sqrt{x+1}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
We can solve the equation a bit easier by letting <math>a = x+1</math>.
 
We can solve the equation a bit easier by letting <math>a = x+1</math>.
Line 31: Line 31:
 
51525 &\bigcirc 51529 \\
 
51525 &\bigcirc 51529 \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
This means that <math>15\sqrt{229}</math> is less than <math>227,</math> so <math>225 + 15\sqrt{229}</math> is less than <math>226</math>.  Thus, the value of the greatest integer less than or equal to <math>x</math> is <math>\boxed{225}</math>.
+
This means that <math>15\sqrt{229}</math> is less than <math>227,</math> so <math>\tfrac{225 + 15\sqrt{229}}{2}</math> is less than <math>226</math>.  Thus, the value of the greatest integer less than or equal to <math>x</math> is <math>\boxed{225}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 19:00, 3 August 2018

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 $\boxed{225}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 54 		Followed by:
Problem 56
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_55&oldid=96744"