How do you solve $x^2 – 4x – 8 = 0$ ?

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Answer 1 · 2016-04-17T17:38:12

$color(green)(x= 2 + sqrt3$

$color(green)(x= 2 - sqrt3)$

$x^2 - 4x -8 =0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=1, b=-4, c= - 8$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (-4)^2-(4 * 1 * -( 8))$

$= 16 + 32 = 48$

The solutions are found using the formula:
$x=(-b+-sqrtDelta)/(2*a)$

$x = (-(-4)+-sqrt(48))/(2*1) = (4+-sqrt(48))/2$

$sqrt48 = sqrt (2*2*2*2*3)= sqrt ( 2^2 *2 ^2 *3)= 4sqrt3$

$x = (4+-sqrt(48))/2 = (4+-4sqrt(3))/2$

$= ((cancel4 ) * ( 1 +- sqrt3)) /cancel2$

$x= (2) * ( 1 + sqrt3)$
$color(green)(x= 2 + sqrt3$

$x= (2) * ( 1 - sqrt3))$
$color(green)(x= 2 - sqrt3)$

How do you solve x^2 – 4x – 8 = 0x^2 – 4x – 8 = 0?