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

(Created page with "== Problem == Let <math>T=\text{TNFTPP}</math>, and let <math>R=T-914</math>. Let <math>x</math> be the smallest real solution of <math>3x^2+Rx+R=90x\sqrt{x+1}</math>. Find the v...")
 
m (→‎Solution)
 
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== Problem ==
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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.''
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Let <math>T=\text{TNFTPP}</math>, and let <math>R=T-914</math>. Let <math>x</math> be the smallest real solution of <math>3x^2+Rx+R=90x\sqrt{x+1}</math>. Find the value of <math>\lfloor x\rfloor</math>.  
 
  
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== Solution ==
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==Problem==
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Let <math>R=675</math>. Let <math>x</math> be the smallest real solution of <math>3x^2+Rx+R=90x\sqrt{x+1}</math>. Find the value of <math>\lfloor x\rfloor</math>.
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==Solution==
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Plugging in <math>R</math> results in
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<cmath>\begin{align*}
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3x^2 + 675x + 675 &= 90x\sqrt{x+1} \\
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x^2 + 225x + 225 &= 30x\sqrt{x+1}
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\end{align*}</cmath>
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We can solve the equation a bit easier by letting <math>a = x+1</math>.
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<cmath>\begin{align*}
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x^2 + 225a &= 30x\sqrt{a} \\
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x^4 + 450x^2 a + 225^2 a^2 &= 900x^2 a \\
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(x^2 - 225a)^2 &= 0 \\
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x^2 - 225x - 225 &= 0
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\end{align*}</cmath>
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Using the [[Quadratic Formula]] yields <math>x = \tfrac{225 \pm 15\sqrt{229}}{2}</math>.  However, we have to check for extraneous solutions and calculate the highest integer less than or equal to <math>x</math>.  This can be easily done using a calculator, but in this solution, we will not use a calculator.
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<br>
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We know that <math>15.1^2 = 228.01</math> and <math>15.15^2 = 229.5225</math>, so <math>226.5 < 15\sqrt{229} < 227.25</math>.  That means <math>-1.5 < \tfrac{225 - 15\sqrt{229}}{2} < -1.125</math>.  Inserting the value into the equation would result in the right hand side having a square root of a negative number, so <math>\tfrac{225 - 15\sqrt{229}}{2}</math> is an extraneous solution.
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<br>
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On the other hand, since we know that there is a solution to the equation and <math>\tfrac{225 + 15\sqrt{229}}{2}</math> makes both sides positive in the original equation, <math>\tfrac{225 + 15\sqrt{229}}{2}</math> is a valid solution.  We know that <math>15\sqrt{229}</math> is greater than <math>225</math>, but we need to know if it’s less than <math>227.</math>
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<cmath>\begin{align*}
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15\sqrt{229} &\bigcirc 227 \\
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225 \cdot 229 &\bigcirc 227^2 \\
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51525 &\bigcirc 51529 \\
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\end{align*}</cmath>
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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>.
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==See Also==
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{{iTest box|year=2007|num-b=54|num-a=56}}
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[[Category:Intermediate Algebra Problems]]

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
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