![]() |
|
Algebra Pre-Algebra 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 $\displaystyle b^2-4ac$ 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 | $\displaystyle \color{blue}{If\ \ ax^2+bx+c=0,\ \ then\ \ x_{1,\ 2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\\;\\ We\ have: \\\;\\k^2+11k+30=0, \\\;\\and \ then \ \ a=1,\ b=11\ \ c=30 . \\\;\\ k_{1,\ 2}= \dfrac{-11\pm\sqrt{11^2-4\cdot30}}{2}=\dfrac{-11\pm\sqrt{121-120}}{2}=\dfrac{-11\pm1}{2} \\\;\\ k_1=\dfrac{-11-1}{2}\ \Longrightarrow\ k_1=-6 \\\;\\ k_2=\dfrac{-11+1}{2}\ \Longrightarrow\ k_2=-5}$ 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 | |
|
![]() | ||||
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 |