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Answer 1 · 2016-10-14T03:27:07

$x_{1} = 4 + 4 \sqrt{3}$ $x_1=4+4 sqrt 3$ and $x_{2} = 4 - 4 \sqrt{3}$ $x_2=4-4 sqrt 3$

$x^{2} - 8 x = 32$ $x^2-8x=32$

$x^{2} - 8 x - 32 = 0$ $x^2-8x-32=0$

I usually use $Delta$ to solve any quadratic equation because I'm not good at guessing :)

$Delta=b^2-4ac$ , with $a=1$ , $b=-8$ and $c=-32$

$Delta=(-8)^2-4(1)(-32)=64+128=192$

$=>sqrt Delta =+-sqrt 192=+-sqrt (2^6xx3)=+-sqrt (2^6) sqrt (3)= +-2^3 sqrt 3$

$=>sqrt Delta=+-8 sqrt 3$

$x_1=(-b+sqrt Delta)/(2a)$ and $x_2=(-b-sqrt Delta)/(2a)$

$x_1=(8+8 sqrt 3)/(2)=(4 color (red) cancel 8(1+1 sqrt 3))/(1 color (red) cancel 2)=4(1+1 sqrt 3)=4+4 sqrt 3$

and

$x_2= (8-8 sqrt 3)/(2)=(4 color (red) cancel8(1-1 sqrt 3))/(1 color (red) cancel 2)=4(1-1 sqrt 3)=4-4 sqrt 3$

Your solution is $x_1=4+4 sqrt 3$ and $x_2=4-4 sqrt 3$

Let me know if you have any question :)

I hope this helps :)