Question #02cfe

1 Answer
Oct 14, 2016

Answer:

#x_1=4+4 sqrt 3# and #x_2=4-4 sqrt 3#

Explanation:

#x^2-8x=32#

#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 :)