How do you solve $x^{2} - 12 x + 5 = 0$ $x^2-12x+5=0$ by completing the square?

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How do you solve $x^{2} - 12 x + 5 = 0$ $x^2-12x+5=0$ by completing the square?

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Answer 1 · 2018-01-15T16:42:16

just add and subtract

Explanation:

so we have
$x^{2} - 12 x + 5 = 0$ $x^2 - 12x + 5 = 0$
$x^{2} - 12 x + 36 - 36 + 5 = 0$ $x^2 - 12x + 36 - 36 + 5 = 0$
${(x + 6)}^{2} - 31 = 0$ $(x+6)^2 -31 = 0$
therefore
$x + 6 = \pm 31^{\frac{1}{2}}$ $x + 6 = +-31^(1/2)$
so
$x = \pm 31^{\frac{1}{2}} - 6$ $x = +-31^(1/2) - 6$
hope u find it helpful :)

Answer 2 · 2018-01-15T18:30:13

$x = 6 \pm \sqrt{31}$ $x=6+-sqrt31$

Explanation:

$using the method of completing the square$ $"using the method of "color(blue)"completing the square"$

$∙ the coefficient of the x^{2} term must be 1 which it is$ $• " the coefficient of the "x^2" term must be 1 which it is"$

$∙ add/subtract {(\frac{1}{2} coefficient of x-term)}^{2} to$ $• " add/subtract "(1/2"coefficient of x-term")^2" to"$
$x^{2} - 12 x$ $x^2-12x$

$\Rightarrow x^{2} + 2 (- 6) x + 36 - 36 + 5 = 0$ $rArrx^2+2(-6)xcolor(red)(+36)color(red)(-36)+5=0$

$\Rightarrow {(x - 6)}^{2} - 31 = 0$ $rArr(x-6)^2-31=0$

$\Rightarrow {(x - 6)}^{2} = 31$ $rArr(x-6)^2=31$

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

$\Rightarrow x - 6 = \pm \sqrt{31} \leftarrow note plus or minus$ $rArrx-6=+-sqrt31larrcolor(blue)"note plus or minus"$

$\Rightarrow x = 6 \pm \sqrt{31}$ $rArrx=6+-sqrt31$

Answer 3 · 2018-01-15T22:38:02

See explanation

Explanation:

For a shortcut method see https://socratic.org/s/aMzZC8RW
This is actually changing

$y = a x^{2} + b x + c$ $y=ax^2+bx+c$ into

$y = a {(x + \frac{b}{2 a})}^{2} + k + c$ $y=a(x+b/(2a))^2+k+c$

Where $a {(\frac{b}{2 a})}^{2} + k = 0$ $a(b/(2a))^2+k=0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$The more formal approach$ $color(blue)("The more formal approach")$

This uses the 'perfect square' but modifies it so that it matches the given equation. A bit like the logic of $something - x + x = something$ $"something"-x+x="something"$

Perfect square $\to {(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $->(a+b)^2=a^2+2ab+b^2$

Compare:

$a^2+2ab+b^2=0" "..."Standard form"$
$x^2-12x+5=0" "..."Given equation"$

$a^2=x^2=>a=x" "......"Point(1)"$

$2ab->2xb->-12x => b=-6" "....."Point(2)"$

Using Point(1) and Point(2) now we have:

$a^2 +2(a)(color(white)("dd")b)color(white)(".")+color(white)("dd")b^2color(white)("dd")=0" Standard form"$
$x^2+2(x)(-6)+(-6)^2=0" Standard form"$
$x^2color(white)("ddd")-12xcolor(white)("dd")+color(white)("dd")5color(white)("d.d")=0" "..."Given equation"$

We now have to change the standard form so that it has the same overall value of the given equation. We need to change $+(-6)^2$ so that it becomes $+5. color(white)("d") ->36-31=+5$ . So the modified standard form is:

$[a^2color(white)()+color(white)("ddd")2abcolor(white)("dd.")+color(white)("dd")b^2color(white)("dd")] -31=0$
$[x^2+2(x)(-6)+(-6)^2]-31=0$

$(a+b)^2-31=0$
$(xcolor(red)(-6))^2color(green)(-31)=0 larr" Vertex form (completing the square)"$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$x_("vertex")=(-1)xx(color(red)(-6)) = +6$
$y_("vertex")=color(green)(-31)$

Vertex $->(x,y)=(6,-31)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For x-intercepts

$x-6=+-sqrt(31)$

$x=6+-sqrt(31)" "$ Exact values

$x=11.57 and x=0.42$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From $y=ax^2+bx+c$ we have:
$y_("intercept")-> c =5$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Tony B

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How do you solve $x^{2} - 12 x + 5 = 0$ $x^2-12x+5=0$ by completing the square?

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How do you solve $x^{2} - 12 x + 5 = 0$ $x^2-12x+5=0$ by completing the square?