How do you solve #x^2+10x=3# by completing the square?
1 Answer
Mar 27, 2016
See explanation...
Explanation:
Add
#x^2+10x+25 = 28#
The left hand side is equal to
#(x+5)^2 = 28 = 2^2*7#
Hence:
#x+5 = +-sqrt(2^2*7) = +-2sqrt(7)#
So:
#x = -5+-2sqrt(7)#
More Info
In general, we find:
#ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))#
From which we can derive the quadratic formula for zeros of
#x = (-b+-sqrt(b^2-4ac))/(2a)#