How do you solve $x^2-2x-4=0$ by factoring?

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Answer 1 · 2016-10-26T12:51:35

$x=1+sqrt5$
OR
$x=1-sqrt5$

The equation can be easily solved by factoring $x^2-2x-4$ which is done by applying the quadratic formula.

$color(blue)(delta=b^2-4ac)=(-2)^2-4(1)(-4)=4+16=20$

$color(green)(x_1=(-b+sqrtdelta)/(2a))=(2+sqrt20)/2=(2+sqrt(4*5))/2=(2+2sqrt5)/2$
$x_1=(1+sqrt5)$

$color(brown)(x_2=(-b-sqrtdelta)/(2a))=(2-sqrt20)/2=(2-sqrt(4*5))/2=(2-2sqrt5)/2$
$x_2=(1-sqrt5)$

$x^2-2x-4=0$
$rArr(x-color(green)(x_1))(x-color(brown)(x_2))=0$

$rArr(x-(1+sqrt5))(x-(1-sqrt5))=0$

$rArr$
$x-(1+sqrt5)=0rArrx=1+sqrt5$
OR
$x-(1-sqrt5)=0rArrx=1-sqrt5$

How do you solve x^2-2x-4=0x^2-2x-4=0 by factoring?