How do you solve using the completing the square method $x^{2} + 2 x + 5 = 0$ $x^ 2 + 2x + 5 = 0$ ?

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Answer 1 · 2016-09-09T15:10:00

$x = - 1 - 2 i$ $x=-1-2i$ or $x = - 1 + 2 i$ $x=-1+2i$

$x^{2} + 2 x + 5 = 0$ $x^2+2x+5=0$

$\Leftrightarrow x^{2} + 2 x + 1 + 4 = 0$ $hArrx^2+2x+1+4=0$

or $x^{2} + 2 \times x \times 1 + 1^{2} - (- 4) = 0$ $x^2+2xx x xx1+1^2-(-4)=0$

or ${(x + 1)}^{2} - {(2 i)}^{2} = 0$ $(x+1)^2-(2i)^2=0$

Now using identity $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ , this becomes

$(x + 1 + 2 i) (x + 1 - 2 i) = 0$ $(x+1+2i)(x+1-2i)=0$

i.e either $x + 1 + 2 i = 0$ $x+1+2i=0$ and $x = - 1 - 2 i$ $x=-1-2i$

or $x + 1 - 2 i = 0$ $x+1-2i=0$ and $x = - 1 + 2 i$ $x=-1+2i$

How do you solve using the completing the square method x2+2x+5=0x^ 2 + 2x + 5 = 0?