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December 21st, 2016, 02:50 AM   #1
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Unhappy SHOW how you solve the equation?

(x+2)^2=36
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December 21st, 2016, 02:52 AM   #2
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$(x+2)^2 = 36 $

$x + 2 = ± \sqrt{36} $

$x = −2 ± 6 $


$x = 4 \;or −8 $
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December 21st, 2016, 02:54 AM   #3
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Quote:
Originally Posted by deesuwalka View Post
$(x+2)^2 = 36 $

$x + 2 = ± \sqrt{36} $

$x = −2 ± 6 $


$x = 4 \;or −8 $
Thank you so much for quick reply
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December 21st, 2016, 05:32 AM   #4
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$$\begin{align*}
(x+2)^2 &= 36 \\
x^2 + 4x + 4 &= 36 \\
x^2 + 4x - 32 &= 0 \\
(x + 8 )(x-4) &= 0 \\
\end{align*}$$
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
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December 21st, 2016, 08:05 AM   #5
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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.
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Last edited by skipjack; December 21st, 2016 at 03:49 PM.
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December 21st, 2016, 05:22 PM   #6
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OK; now let's see your stuff:

(2x+3)^2=81
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December 22nd, 2016, 07:28 PM   #7
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Quote:
Originally Posted by Denis View Post
OK; now let's see your stuff:

(2x+3)^2=81
x=3 or x=-6
Am I right?
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December 22nd, 2016, 07:44 PM   #8
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Yes!!
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December 23rd, 2016, 11:04 AM   #9
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Quote:
Originally Posted by Denis View Post
OK; now let's see your stuff:

(2x+3)^2=81

$\displaystyle \it \color{blue} {(2x+3)^2 = 81\Leftrightarrow \sqrt{(2x+3)^2}=\sqrt{81}\Leftrightarrow |2x+3| = 9 \Leftrightarrow 2x+3 = \pm9\Leftrightarrow
\\\;\\
\Leftrightarrow 2x+3\in\{-9,\ 9\}|_{-3} \Leftrightarrow 2x\in\{-12,\ 6\}|_{:2}\Leftrightarrow x\in\{-6, \ 3\}}$
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December 23rd, 2016, 11:07 AM   #10
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Aurel, that'll scare poor Davil away !
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