How do you solve $x^{2} + 22 x + 121 = 0$ $x^2+22x+121=0$ ?

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How do you solve $x^{2} + 22 x + 121 = 0$ $x^2+22x+121=0$ ?

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Answer 1 · 2018-06-23T12:42:07

$x = - 11 (multiplicity 2)$ $x=-11 " (multiplicity 2)"$

Explanation:

$this is a perfect square$ $"this is a "color(blue)"perfect square"$

$∙ x {(a \pm b)}^{2} = a^{2} \pm 2 a b + b^{2}$ $•color(white)(x)(a+-b)^2=a^2+-2ab+b^2$

$with a = x and b = 11$ $"with "a=x" and "b=11$

$2 a b = 2 \times x \times 11 = 22 x$ $2ab=2xx x xx11=22x$

${(x + 11)}^{2} = 0$ $(x+11)^2=0$

$(x + 11) (x + 11) = 0$ $(x+11)(x+11)=0$

$\Rightarrow x = - 11 of multiplicity 2$ $rArrx=-11" of multiplicity 2"$

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How do you solve $x^{2} + 22 x + 121 = 0$ $x^2+22x+121=0$ ?

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How do you solve $x^{2} + 22 x + 121 = 0$ $x^2+22x+121=0$ ?