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July 17th, 2014, 07:52 AM   #1
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Year 11 Completing the Square Question

Question:

x^2 + 6x = -11

How would I solve for x?


The answer at the back of the book is:
x = -3 +/- √11

Help me please! Its for my maths homework.

I believe the solution is found by using completing the square because the example above the question uses this method:
The example was:
x^2 + 4x + 1 = 0
the answer is: x = -2 +/- √3

Last edited by wrightarya; July 17th, 2014 at 08:03 AM. Reason: additional info that could be helpful to solve problem
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July 17th, 2014, 09:05 AM   #2
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Quote:
Originally Posted by wrightarya View Post
Question:

x^2 + 6x = -11

How would I solve for x?

$x^2+6x+11=0
\\\;\\
x_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}
\\\;\\
\Delta = b^2-4ac.
\\\;\\
To \ further, \ with \ care\ caution, \ it\ is \ dry\ sand
\\\;\\



(Analog,\ for
\\\;\\
\ x^2\ +\ 6x\ -\ 11\ =\ 0\ \ ! \ )$
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July 17th, 2014, 10:49 AM   #3
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Quote:
Originally Posted by wrightarya View Post
x^2 + 6x = -11
Pls check; shouldn't that be: x^2 + 6x = 2 ?
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July 17th, 2014, 07:31 PM   #4
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I triple checked the question and answer in the textbook. I haven't wrote it down incorrectly.

I substituted the answer in for x and I see how it the question should be
x^ + 6x = 2

maybe the Haese year 11 textbook made a misprint!
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July 18th, 2014, 12:43 AM   #5
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The solution for the example question is:

x^2 + 4x + 1 = 0

x^2 + 4x + 2^2 + 1 - 2^2 = 0
(x+2)^2 -3 = 0
(x+2)^2 = 3
x+2 = √3
x= -2 +/- √3

That's how they solved it in the book. But this method doesn't seem to work for the question I posted which was this question:
x^2 + 6x = -11

Last edited by wrightarya; July 18th, 2014 at 01:08 AM.
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July 18th, 2014, 12:57 AM   #6
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Quote:
Originally Posted by wrightarya View Post
(x+2)^2 = 3
x+2 = √3
x= -2 +/- √3

Does that help?
$\Large (x+2)^2 = 3\ \Longleftrightarrow\ \sqrt{ (x+2)^2}=\sqrt3 \Longleftrightarrow\ |x+2 |= \sqrt 3\Longleftrightarrow\ x+2 = \pm\sqrt 3$
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July 18th, 2014, 01:13 AM   #7
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Quote:
Originally Posted by wrightarya View Post
Question:

x^2 + 6x = -11

How would I solve for x?


The answer at the back of the book is:
x = -3 +/- √11
Remark

If $x^2+bx+c=0$

and $\displaystyle x_1,\ x_2$ are the solutions (roots),

then $\displaystyle x^2+bx+c=(x-x_1)(x-x_2).$
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