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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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 · 2016-02-25T06:55:47

$(1 - \sqrt{6})$ $(1-sqrt6)$ and $(1 + \sqrt{6})$ $(1+sqrt6)$

Explanation:

To solve $x^{2} - 2 x - 5 = 0$ $x^2−2x−5=0$ using the completing the square method, one picks up terms containing $x^{2}$ $x^2$ and $x$ $x$ . Here these are $x^{2} - 2 x$ $x^2−2x$ . Here coefficient of $x^{2}$ $x^2$ is one and that of $x$ $x$ is $- 2$ $-2$ , hence adding half of the square of latter i.e. adding ${(- 1)}^{2}$ $(-1)^2$ (i..e. $1$ $1$ ) should make $x^{2} - 2 x$ $x^2−2x$ , a complete square.

Hence for solving, let us add and subtract $1$ $1$ , and doing this the equation becomes

$x^{2} - 2 x + 1 - 6 = 0$ $x^2−2x+1−6=0$

or ${(x - 1)}^{2} - {(\sqrt{6})}^{2} = 0$ $(x-1)^2-(sqrt6)^2=0$

i.e. $(x - 1 + \sqrt{6}) (x - 1 - \sqrt{6}) = 0$ $(x-1+sqrt6)(x-1-sqrt6)=0$

i.e. $x = (1 - \sqrt{6})$ $x=(1-sqrt6)$ or $x = (1 + \sqrt{6})$ $x=(1+sqrt6)$

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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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Explanation:

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Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

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