How do you solve using the completing the square method $x^{2} + 12 x + 20 = 0$ $x^2 + 12x + 20 = 0$ ?

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Answer 1 · 2016-03-27T17:05:57

For an in depth explanation of method see
http://socratic.org/s/at7MQHg2

$x \approx - 1.101 x \approx - 10.899$ $color(blue)(x~~-1.101" "x~~-10.899)$

Using Shortcuts

Given: $y = x^{2} + 12 x + 12$ $" "y=x^2+12x+12$

Let $k$ $k$ be the error adjustment constant

$y = x^{2} + 12 x + 12 + k$ $y=x^2+12x+12+k$

Step 1: $y = (x^{2} + 12 x) + 12 + k$ $" "y=(x^color(magenta)(2)+12x)+12+k$

Step 2: $y = {(x + 12 x)}^{2} + 12 + k$ $" "y=(x+color(green)(12)x)^(color(magenta)(2))+12+k$

Step 3: $use (\frac{1}{2}) \times (12) = 6$ $" use "(1/2)xx(color(green)(12))=color(magenta)(6)$

$\Rightarrow y = {(x + 6)}^{2} + 12 + k$ $=> y=(xcolor(magenta)(+6))^2+12+k$

Step 4: ${(+ 6)}^{2} + k = 0$ $" "color(magenta)((+6))^2+k=0$

$\Rightarrow k = - 36$ $=>k=color(green)(-36)$

$\Rightarrow y = {(x + 6)}^{2} + 12 - 36$ $=>y=(x+6)^2+12-36$

$\Rightarrow y = {(x + 6)}^{2} - 24$ $=>y=(x+6)^2-24$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~

At $y = 0; x = \pm \sqrt{24} - 6$ $y=0"; " x=+- sqrt(24)-6$

$x \approx - 1.101 x \approx - 10.899$ $color(blue)(x~~-1.101" "x~~-10.899)$

$Notice that plots for both equations match$ $color(magenta)("Notice that plots for both equations match")$
Tony B

How do you solve using the completing the square method x2+12x+20=0x^2 + 12x + 20 = 0?