How do you solve $x^2 - 2x = 32$ ?

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Answer 1 · 2017-04-08T10:59:10

Solutuion : $x= 1 + sqrt 33 or x= 1 - sqrt 33$

$x^2-2x=32 or x^2-2x +1 =32 +1 or (x -1)^2 =33 or x -1 = +- sqrt 33$

or $x= 1 + sqrt 33 or x = 1 - sqrt 33$

Solutuion : $x= 1 + sqrt 33 or x= 1 - sqrt 33$ [Ans]

Answer 2 · 2017-04-08T11:36:12

$color(blue)(x=1-sqrt33$ or $color(blue)(x=1+sqrt33$

$x^2-2x=32$

$:.x^2-2x-32=0$

$:.x=(-b+-sqrt(b^2-4ac))/(2a)$

$:.x=-(-2)+-sqrt((-2)^2-4(1)(-32))/(2(1)$

$:.x=(2+-sqrt(4+128))/2$

$:.x=(2+-sqrt132)/2$

$:.x=2+-sqrt(2*2*3*11)/2$

$:.x=(2+-2sqrt33)/2$

$:.x=(2-2sqrt33)/2$

$:.x=(cancel2^color(blue)1(1-sqrt33))/cancel2^color(blue)1$

$:.x=1-sqrt33$

or:

$:.x=(2+2sqrt33)/2$

$:.x=cancel2^color(blue)1(1+sqrt33)/cancel2^color(blue)1$

$:.color(blue)(x=1+sqrt33$

How do you solve x^2 - 2x = 32x^2 - 2x = 32?