How do you solve $x^{2} - 4 x - 9 = 0$ $x^2 - 4x - 9 = 0$ by completing the square?

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How do you solve $x^{2} - 4 x - 9 = 0$ $x^2 - 4x - 9 = 0$ by completing the square?

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Answer 1 · 2017-06-06T17:31:41

$x = 2 \pm \sqrt{13}$ $x=2+-sqrt13$

Explanation:

$to complete the square$ $"to complete the square"$

add ${(\frac{1}{2} coefficient of x-term)}^{2} to both sides$ $(1/2"coefficient of x-term")^2 " to both sides"$

$that is add {(- \frac{4}{2})}^{2} = 4 to both sides$ $"that is add " (-4/2)^2=4" to both sides"$

$\Rightarrow x^{2} - 4 x + 4 - 9 = 0 + 4$ $rArrx^2-4xcolor(red)(+4)-9=0color(red)(+4)$

$\Rightarrow {(x - 2)}^{2} - 9 = 4$ $rArr(x-2)^2-9=4$

$\Rightarrow {(x - 2)}^{2} = 13$ $rArr(x-2)^2=13$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(x - 2)}^{2}} = \pm \sqrt{13} \leftarrow note plus or minus$ $sqrt((x-2)^2)=+-sqrt13larr" note plus or minus"$

$\Rightarrow x - 2 = \pm \sqrt{13}$ $rArrx-2=+-sqrt13$

$\Rightarrow x = 2 \pm \sqrt{13}$ $rArrx=2+-sqrt13$

$or x \approx 5.6, x \approx - 1.6 to 1 decimal place$ $"or " x~~5.6, x~~-1.6" to 1 decimal place"$

Answer 2 · 2017-06-06T18:10:24

$x = + 5.606 or x = - 1.606$ $x = +5.606 or x= -1.606$

Explanation:

$x^{2} - 4 x - 9 = 0$ $x^2 -4x -9=0$

To complete the square means exactly what it says...

"make an expression into a perfect square by adding the part that is missing..."

The steps are given, refer to the details of the working below:

In $a x^{2} + b x + c = 0$ $ax^2 +bx +c =0$

$1 : \to 1 x^{2} - 4 x - 9 = 0$ $1:rarr" "1x^2-4x" "color(red)(-9)=0$

$2 : \to x^{2} - 4 x = 9$ $2:rarr" "x^2 color(blue)(-4)x" " = color(red)(9)$

$3 : \to x^{2} - 4 x + 4 = 9 + 4$ $3:rarr" "x^2 -4x " " color(blue)(+4) = 9 color(blue)( +4)$

$4 : \to {(x - 2)}^{2} = 13$ $4:rarr" "(x-2)^2" " = 13$

$5 : \to x - 2 = \pm \sqrt{13}$ $5:rarr" "x-2" " = +-sqrt13$

$6 : \to x w w w w w w w = + \sqrt{13} + 2 = + 5.606$ $6:rarr" "xcolor(white)(wwwwwww) = +sqrt13 +2 = +5.606$
$6 : \to x w w w w w w w = - \sqrt{13} + 2 = - 1.606$ $6:rarr" "xcolor(white)(wwwwwww) = -sqrt13 +2 = -1.606$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Step 1: Make $a = 1 (a$ $a=1" " (a$ is already equal to $1$ $1$ )
Step 2: Move the constant to the other side.
Step 3: complete the square by adding ${(\frac{b}{2})}^{2}$ $color(blue)((b/2)^2)$ to both sides.
In this case $b = - 4$ $color(blue)(b=-4)" "$ so, ${(\frac{- 4}{2})}^{2} = {(- 2)}^{2} = + 4$ $color(blue)(((-4)/2)^2 = (-2)^2 =+4)$
Step 4: Write the LHS as the square of a binomial
Step 5: square root both sides $\to$ $rarr$ remember $\pm \sqrt{}$ $+-sqrt$
Step 6: solve for $x$ $x$ to get 2 values

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How do you solve $x^{2} - 4 x - 9 = 0$ $x^2 - 4x - 9 = 0$ by completing the square?

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How do you solve $x^{2} - 4 x - 9 = 0$ $x^2 - 4x - 9 = 0$ by completing the square?