How do you solve using the completing the square method $x^{2} + 2 x - 15 = 0$ $x^2 + 2x - 15 = 0$ ?

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How do you solve using the completing the square method $x^{2} + 2 x - 15 = 0$ $x^2 + 2x - 15 = 0$ ?

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Answer 1 · 2016-04-04T16:41:14

The solutions are:
$x = 3$ $color(green)(x =3$ or $x = - 5$ $color(green)(x = -5$

Explanation:

$x^{2} + 2 x - 15 = 0$ $x^2 + 2x - 15 =0$

$x^{2} + 2 x = 15$ $x^2 + 2x = 15$

To write the Left Hand Side as a Perfect Square, we add 1 to both sides:

$x^{2} + 2 x + 1 = 15 + 1$ $x^2 + 2x + 1 = 15 + 1$

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 16$ $x^2 + 2 * x *1 + 1^2 = 16$

Using the Identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get

${(x + 1)}^{2} = 16$ $(x+1)^2 = 16$

$x + 1 = \sqrt{16}$ $x + 1 = sqrt16$ or $x + 1 = - \sqrt{16}$ $x + 1 = -sqrt16$

$x = 4 - 1 = 3$ $color(green)(x = 4 - 1 =3$ or $x = 4 - 1 = - 5$ $color(green)(x = 4 - 1 = -5$

Answer 2 · 2016-04-04T17:05:56

$\Rightarrow + x = - 1 \pm 4$ $color(blue)(=> +x=-1+-4)$

$x = + 3 or - 5$ $color(brown)(x= +3 " or "-5)$

Explanation:

The equation in standard form is:
$y = a x^{2} + b x + c$ $" "color(green)(y=ax^2+bx+c)$

The equation in vertex form is:
$y = a {(x + \frac{b}{2 a})}^{2} + c - [{(\frac{b}{2})}^{2}]$ $" "color(blue)(y=a(x+b/(2a))^2+c-[(b/2)^2])$

$The - [{(\frac{b}{2})}^{2}] corrects the error produced by {(\frac{b}{2 a})}^{2}$ $color(brown)("The "-[(b/2)^2]" corrects the error produced by" (b/(2a))^2$

They are different versions of the same thing!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$Solving your question$ $color(blue)("Solving your question")$

$a = 1; b = 2; c = - 15$ $a = 1" ; "b=2" ; "c=-15$ giving

$y = {(x + 1)}^{2} - 15 - 1$ $y= (x+1)^2-15-1$

$y = {(x + 1)}^{2} - 16$ $color(blue)(y=(x+1)^2-16)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set the vertex form equation as equal to zero

$\Rightarrow 0 = {(x + 1)}^{2} - 16$ $=>0=(x+1)^2-16$

$\Rightarrow {(x + 1)}^{2} = + 16$ $=> (x+1)^2=+16$

Taking the square root of both sides

$\Rightarrow \pm (x + 1) = \pm \sqrt{16}$ $=> +-(x+1)=+-sqrt(16)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Consider the case - (x + 1)$ $color(blue)("Consider the case "-(x+1))$

$\Rightarrow - x - 1 = \pm 4$ $=> -x-1=+-4$

$\Rightarrow - x = + 1 \pm 4$ $=>-x=+1+-4$

Multiply both sides by (-1)

$\Rightarrow + x = - 1 \pm 4$ $color(blue)(=> +x=-1+-4)$

$x = + 3 or - 5$ $color(brown)(x= +3 " or "-5)$

'....................................................................

$Consider the case + (x + 1)$ $color(blue)("Consider the case "+(x+1))$

$\Rightarrow x = - 1 \pm 4$ $=>x=-1+-4$

$Same as for - (x + 1)$ $color(brown)("Same as for "-(x+1))$
'...................................................

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How do you solve using the completing the square method $x^{2} + 2 x - 15 = 0$ $x^2 + 2x - 15 = 0$ ?

2 Answers

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How do you solve using the completing the square method x2+2x−15=0x^2 + 2x - 15 = 0?

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How do you solve using the completing the square method $x^{2} + 2 x - 15 = 0$ $x^2 + 2x - 15 = 0$ ?