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 $ 
December 21st, 2016, 02:54 AM  #3 
Newbie Joined: Dec 2016 From: Chicago Posts: 2 Thanks: 0  
December 21st, 2016, 05:32 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,557 Thanks: 2148 Math Focus: Mainly analysis and algebra 
$$\begin{align*} (x+2)^2 &= 36 \\ x^2 + 4x + 4 &= 36 \\ x^2 + 4x  32 &= 0 \\ (x + 8 )(x4) &= 0 \\ \end{align*}$$ And because the product is zero, at least one of the factors is zero. Thus $$\left. \begin{aligned} x+8 &= 0 \\ x &= 8 \end{aligned} \right\} \; \text{or} \; \left\{ \begin{aligned} x4 &= 0 \\ x &= 4 \end{aligned} \right.$$ ("Or" because it's not possible that both are true simultaneously). I'll leave you to demonstrate the factorisation of the cuadratic 
December 21st, 2016, 08:05 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,354 Thanks: 591 
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. 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: 8,788 Thanks: 607 
OK; now let's see your stuff: (2x+3)^2=81 
December 22nd, 2016, 07:28 PM  #7 
Newbie Joined: Dec 2016 From: Philippines Posts: 2 Thanks: 3 Math Focus: Algebra  
December 22nd, 2016, 07:44 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,788 Thanks: 607 
Yes!!

December 23rd, 2016, 11:04 AM  #9 
Senior Member Joined: Apr 2014 From: Europa Posts: 570 Thanks: 174  $\displaystyle \it \color{blue} {(2x+3)^2 = 81\Leftrightarrow \sqrt{(2x+3)^2}=\sqrt{81}\Leftrightarrow 2x+3 = 9 \Leftrightarrow 2x+3 = \pm9\Leftrightarrow \\\;\\ \Leftrightarrow 2x+3\in\{9,\ 9\}_{3} \Leftrightarrow 2x\in\{12,\ 6\}_{:2}\Leftrightarrow x\in\{6, \ 3\}}$ 
December 23rd, 2016, 11:07 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,788 Thanks: 607 
Aurel, that'll scare poor Davil away !


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