
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 18th, 2014, 03:23 AM  #1 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  Quadratic formula
The question was $\displaystyle k^2 +11k+30=0$ solve k. My answer $\displaystyle k = \frac{11± \sqrt{2}}{2}$ The books answer is $\displaystyle k = \frac{11± 1}{2}$ If we take a step back from the final solution I was left with $\displaystyle k \frac{11}{2}= \sqrt{ \frac{2}{4}}=\frac{ \sqrt{2}}{2}$ Can someone show or explain in words what step I missed? Last edited by skipjack; November 22nd, 2014 at 03:47 AM. 
October 18th, 2014, 03:53 AM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 
You simply didn't calculate the discriminant $\displaystyle b^24ac$ correctly.

October 18th, 2014, 06:14 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
\begin{align*} k^2 + 11k + 30 &= 0 \\ k^2 + 2\tfrac{11}{2}k + \left( \tfrac{11}{2} \right)^2  \left( \tfrac{11}{2} \right)^2 + 30 &= 0 \\ \left( k + \tfrac{11}{2} \right)^2  \tfrac{121}{4} + \tfrac{120}{4} &= 0 \\ \left( k + \tfrac{11}{2} \right)^2 &= \tfrac{121}{4}  \tfrac{120}{4} = \tfrac14 \\ k + \tfrac{11}{2} &= \pm \tfrac12 \end{align*}

October 18th, 2014, 06:17 AM  #4 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  $\displaystyle \color{blue}{If\ \ ax^2+bx+c=0,\ \ then\ \ x_{1,\ 2}=\dfrac{b\pm\sqrt{b^24ac}}{2a} \\\;\\ We\ have: \\\;\\k^2+11k+30=0, \\\;\\and \ then \ \ a=1,\ b=11\ \ c=30 . \\\;\\ k_{1,\ 2}= \dfrac{11\pm\sqrt{11^24\cdot30}}{2}=\dfrac{11\pm\sqrt{121120}}{2}=\dfrac{11\pm1}{2} \\\;\\ k_1=\dfrac{111}{2}\ \Longrightarrow\ k_1=6 \\\;\\ k_2=\dfrac{11+1}{2}\ \Longrightarrow\ k_2=5}$ Last edited by aurel5; October 18th, 2014 at 06:23 AM. 
October 21st, 2014, 01:01 AM  #5  
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  Quote:
ohhhhhhhhhhhhhhhh. $\displaystyle 11 \cdot 11 = 121$ Edit: the book lists k=5 or 6, but............. Aren't we subtracting 11/2 here? K+11/2=±1/2 Wouldn't it be 11/2 from +1/2 =10/2 = 5 11/2 from 1/2 =12/2 = 6 ???? Last edited by Opposite; October 21st, 2014 at 01:09 AM.  
November 22nd, 2014, 03:51 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,288 Thanks: 1968  
November 22nd, 2014, 12:16 PM  #7  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
because 2 isn't being multiplied by $\displaystyle \ \tfrac{11}{2}k.$ $\displaystyle k^2 + 2\left(\tfrac{11}{2}k \right) + \left( \tfrac{11}{2} \right)^2  \left( \tfrac{11}{2} \right)^2 + 30 = 0 $ $\displaystyle 2\tfrac{11}{2}, \ \ like \ \ any \ \ other \ \ mixed \ \ number \ \ in \ \ general, \ \ here \ \ specifically \ \ equals \ \ \tfrac{15}{2}.$  
November 22nd, 2014, 04:17 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
I don't know of anybody who would consider that to be a mixed number. If you were using mixed numbers you'd write $5 \frac12$ rather than $\frac{11}{2}$.


Tags 
formula, quadratic 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Quadratic Formula  mathkid  Algebra  15  September 30th, 2012 11:01 PM 
Quadratic Formula Help?  wrongnmbr  Algebra  7  August 29th, 2010 09:25 PM 
Using the Quadratic Formula???  confused20  Algebra  3  March 10th, 2008 05:34 PM 
NEXT? quadratic formula  Sean  Algebra  1  December 28th, 2006 04:39 PM 
Quadratic Formula  mathkid  Calculus  0  December 31st, 1969 04:00 PM 