My Math Forum Year 11 Completing the Square Question
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 17th, 2014, 07:52 AM #1 Newbie   Joined: Jul 2014 From: Australia Posts: 24 Thanks: 0 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
July 17th, 2014, 09:05 AM   #2
Senior Member

Joined: Apr 2014
From: Europa

Posts: 575
Thanks: 176

Quote:
 Originally Posted by wrightarya Question: x^2 + 6x = -11 How would I solve for x?
$x^2+6x+11=0 \\\;\\ x_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a} \\\;\\ \Delta = b^2-4ac. \\\;\\ To \ further, \ with \ care\ caution, \ it\ is \ dry\ sand \\\;\\ (Analog,\ for \\\;\\ \ x^2\ +\ 6x\ -\ 11\ =\ 0\ \ ! \ )$

July 17th, 2014, 10:49 AM   #3
Math Team

Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 13,476
Thanks: 950

Quote:
 Originally Posted by wrightarya x^2 + 6x = -11
Pls check; shouldn't that be: x^2 + 6x = 2 ?

 July 17th, 2014, 07:31 PM #4 Newbie   Joined: Jul 2014 From: Australia Posts: 24 Thanks: 0 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!
 July 18th, 2014, 12:43 AM #5 Newbie   Joined: Jul 2014 From: Australia Posts: 24 Thanks: 0 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.
July 18th, 2014, 12:57 AM   #6
Senior Member

Joined: Apr 2014
From: Europa

Posts: 575
Thanks: 176

Quote:
 Originally Posted by wrightarya (x+2)^2 = 3 x+2 = √3 x= -2 +/- √3 Does that help?
$\Large (x+2)^2 = 3\ \Longleftrightarrow\ \sqrt{ (x+2)^2}=\sqrt3 \Longleftrightarrow\ |x+2 |= \sqrt 3\Longleftrightarrow\ x+2 = \pm\sqrt 3$

July 18th, 2014, 01:13 AM   #7
Senior Member

Joined: Apr 2014
From: Europa

Posts: 575
Thanks: 176

Quote:
 Originally Posted by wrightarya 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 $\displaystyle x^2+bx+c=(x-x_1)(x-x_2).$

 Tags algebra, completing, quadratic, question, square, year, year 11

« max | complex »

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post wrightarya New Users 6 July 18th, 2014 02:18 AM GuyDoingMathFoiled Algebra 10 January 9th, 2014 04:24 PM link107 Algebra 2 September 8th, 2012 03:42 PM AlwaysALilLost Algebra 6 May 15th, 2007 02:04 PM morning_mood Algebra 10 May 4th, 2007 07:29 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top