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October 18th, 2014, 03:23 AM   #1
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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.
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October 18th, 2014, 03:53 AM   #2
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You simply didn't calculate the discriminant $\displaystyle b^2-4ac$ correctly.
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October 18th, 2014, 06:14 AM   #3
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\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*}
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October 18th, 2014, 06:17 AM   #4
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Quote:
Originally Posted by Opposite View Post
$\displaystyle k^2 +11k+30=0$ solve k.

$\displaystyle \color{blue}{If\ \ ax^2+bx+c=0,\ \ then\ \ x_{1,\ 2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
\\\;\\
We\ have:
\\\;\\k^2+11k+30=0,
\\\;\\and \ then \ \ a=1,\ b=11\ \ c=30 .
\\\;\\
k_{1,\ 2}= \dfrac{-11\pm\sqrt{11^2-4\cdot30}}{2}=\dfrac{-11\pm\sqrt{121-120}}{2}=\dfrac{-11\pm1}{2}
\\\;\\
k_1=\dfrac{-11-1}{2}\ \Longrightarrow\ k_1=-6
\\\;\\
k_2=\dfrac{-11+1}{2}\ \Longrightarrow\ k_2=-5}$
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Last edited by aurel5; October 18th, 2014 at 06:23 AM.
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October 21st, 2014, 01:01 AM   #5
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Quote:
Originally Posted by Pero View Post
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.
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November 22nd, 2014, 03:51 AM   #6
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Quote:
Originally Posted by Opposite View Post
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.
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November 22nd, 2014, 12:16 PM   #7
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Quote:
Originally Posted by v8archie View Post
$\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 $


$\displaystyle 2\tfrac{11}{2}, \ \ like \ \ any \ \ other \ \ mixed \ \ number \ \ in \ \ general, \ \ here \ \ specifically \ \ equals \ \ \tfrac{15}{2}.$
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November 22nd, 2014, 04:17 PM   #8
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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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