How do you solve #x(x - 4) = 45 # by completing the square?

1 Answer
May 23, 2016

Answer:

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