How do you solve the equation $x^{2} - 2 x - 10 = 0$ $x^2-2x-10=0$ by completing the square?

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Answer 1 · 2017-06-06T17:46:44

$x = 1 \pm \sqrt{11}$ $x=1+-sqrt11$

$x^{2} - 2 x - 10 = 0$ $x^2-2x-10=0$

half the coefficient of $x$ $x$ square it, add and subtract to the equation

$(x^{2} - 2 x + 1^{2}) - 1^{2} - 10 = 0$ $(x^2-2x+1^2)-1^2-10=0$

The bracket is a perfect square

$\Rightarrow {(x - 1)}^{2} - 11 = 0$ $=>(x-1)^2-11=0$

${(x - 1)}^{2} = 11$ $(x-1)^2=11$

$x - 1 = \pm \sqrt{11}$ $x-1=+-sqrt11$

$x = 1 \pm \sqrt{11}$ $x=1+-sqrt11$

How do you solve the equation x2−2x−10=0x^2-2x-10=0 by completing the square?