How do you solve #8h ^ { 2} - 8h + 1= 0#?

1 Answer
Morgan
May 12, 2017

Explanation:

You can use the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

to solve equations of the form #ax^2+bx+c=0#. In this case, #a=8,b=-8,# and #c=1#.

#x=(-(-8)+-sqrt((-8)^2-4(8)(1)))/(2(8))#

#=(8+-sqrt(32))/16#

#=(8+-sqrt(16*2))/16#

#=(8+-4sqrt(2))/16#

#=1/2+-sqrt(2)/4#

Therefore, #x=1/2+sqrt(2)/4# or #x=1/2-sqrt(2)/4# are solutions.

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