How do you solve #x(x - 4) = 45 # by completing the square?
1 Answer
May 23, 2016
x = -5 , x= 9
Explanation:
Since this is a quadratic equation expand brackets and equate to zero.
#rArrx^2-4x-45=0# This is now in standard form :
#ax^2+bx+c=0# To complete the square add on
#(b/2)^2# here b = -4
#rArr(-4/2)^2=4# equation can now be written as
#[x^2-4x+4]+(-4)-45=0# Since we added on 4 to complete the square we must -4
#rArr(x-2)^2-4-45=0rArr(x-2)^2=49# Taking the square root of both sides.
#x-2=±sqrt49=±7# hence x = 7 + 2 = 9 or x = -7 + 2 =-5
