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 December 21st, 2016, 02:50 AM #1 Newbie   Joined: Dec 2016 From: Chicago Posts: 2 Thanks: 0 SHOW how you solve the equation? (x+2)^2=36
 December 21st, 2016, 02:52 AM #2 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $(x+2)^2 = 36$ $x + 2 = ± \sqrt{36}$ $x = −2 ± 6$ $x = 4 \;or −8$ Thanks from Davil and astro511
December 21st, 2016, 02:54 AM   #3
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Quote:
 Originally Posted by deesuwalka $(x+2)^2 = 36$ $x + 2 = ± \sqrt{36}$ $x = −2 ± 6$ $x = 4 \;or −8$
Thank you so much for quick reply

 December 21st, 2016, 05:32 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 Math Focus: Mainly analysis and algebra \begin{align*} (x+2)^2 &= 36 \\ x^2 + 4x + 4 &= 36 \\ x^2 + 4x - 32 &= 0 \\ (x + 8 )(x-4) &= 0 \\ \end{align*} And because the product is zero, at least one of the factors is zero. Thus \left. \begin{aligned} x+8 &= 0 \\ x &= -8 \end{aligned} \right\} \; \text{or} \; \left\{ \begin{aligned} x-4 &= 0 \\ x &= 4 \end{aligned} \right. ("Or" because it's not possible that both are true simultaneously). I'll leave you to demonstrate the factorisation of the cuadratic Thanks from Davil
 December 21st, 2016, 08:05 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,878 Thanks: 766 Deesuwalka's method is simpler though V8Archie's is more general. Notice that if you were given a value for x and asked to evaluate $\displaystyle (x+2)^2$ you would 1) add 2 to x 2) square Here, you are given the value, 36, and asked to find the x that will give that value. So you need to "back out": do the opposite- instead of "add 2" you "subtract 2" and instead of "squaring" take the square root. And, of course, you have to reverse the order- first take the square root, then add 2 as Deesuwalka did. Thanks from Davil Last edited by skipjack; December 21st, 2016 at 03:49 PM.
 December 21st, 2016, 05:22 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,082 Thanks: 723 OK; now let's see your stuff: (2x+3)^2=81 Thanks from Joppy and Davil
December 22nd, 2016, 07:28 PM   #7
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Math Focus: Algebra
Quote:
 Originally Posted by Denis OK; now let's see your stuff: (2x+3)^2=81
x=3 or x=-6
Am I right?

 December 22nd, 2016, 07:44 PM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,082 Thanks: 723 Yes!! Thanks from Joppy and Davil
December 23rd, 2016, 11:04 AM   #9
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Joined: Apr 2014
From: Europa

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Quote:
 Originally Posted by Denis OK; now let's see your stuff: (2x+3)^2=81

$\displaystyle \it \color{blue} {(2x+3)^2 = 81\Leftrightarrow \sqrt{(2x+3)^2}=\sqrt{81}\Leftrightarrow |2x+3| = 9 \Leftrightarrow 2x+3 = \pm9\Leftrightarrow \\\;\\ \Leftrightarrow 2x+3\in\{-9,\ 9\}|_{-3} \Leftrightarrow 2x\in\{-12,\ 6\}|_{:2}\Leftrightarrow x\in\{-6, \ 3\}}$

 December 23rd, 2016, 11:07 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,082 Thanks: 723 Aurel, that'll scare poor Davil away !

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