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 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. Thanks from topsquark and Opposite
 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*} Thanks from topsquark
October 18th, 2014, 06:17 AM   #4
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Quote:
 Originally Posted by Opposite $\displaystyle k^2 +11k+30=0$ solve k.

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

Last edited by aurel5; October 18th, 2014 at 06:23 AM.

October 21st, 2014, 01:01 AM   #5
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Quote:
 Originally Posted by Pero You simply didn't calculate the discriminant $\displaystyle b^2-4ac$ correctly.
$\displaystyle 11 \cdot 11 \neq 122$

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
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Quote:
 Originally Posted by Opposite The question was $\displaystyle k^2 +11k+30=0$
Are you sure the book didn't give $\displaystyle k^2 - 11k + 30 = 0$? Solving that gives $k$ = 5 or 6.

November 22nd, 2014, 12:16 PM   #7
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Quote:
 Originally Posted by v8archie $\displaystyle k^2 + 2\tfrac{11}{2}k + \left( \tfrac{11}{2} \right)^2 - \left( \tfrac{11}{2} \right)^2 + 30 = 0$
Grouping symbols are required where I show some added below,
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 $\frac{11}{2}$.

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