How do you solve #x^2-4.7x=-2.8# by completing the square?

2 Answers
Apr 23, 2017

Answer:

#color(red)(x=4# or #color(red)(x=0.7#

Explanation:

Completing the square:

#:.x^2-4.7x+2.8=0#

#:.x^2-4.7x+(-4.7/2)^2=-2.8+(-4.7/2)^2#

#:.x^2-4.7x+5.5225=-2.8+5.5225#

#:.(x-2.35)^2=2.7225#

#:.sqrt((x-2.35)^2)=sqrt(2.7225)#

#:.x-2.35=+-1.65#

#:.x=2.35+-1.65#

#:.color(red)(x=2.35+1.65=4#

or #:.color(red)(x=2.35-1.65=0.7#

check:

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

#:.x=(-(-4.7)+-sqrt((-4.7)^2-4(1)(2.8)))/(2a)#

#:.x=(-(-4.7)+-sqrt(22.09-11.2))/(2a)#

#:.x=(4.7+-sqrt10.89)/2#

#:.x=(4.7+-3.3)/2#

#:.x=(4.7+3.3)/2#

#:.color(red)(x=8/2=4#

or # :. x=(4.7-3.3)/2#

#:.color(red)(x=1.4/2=0.7#

Apr 23, 2017

Answer:

#x=4" "# or #" "x=0.7#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(20x-47)# and #b=33# below.

Given:

#x^2-4.7x=-2.8#

Add #2.8# to both sides to get:

#x^2-4.7x+2.8 = 0#

Multiply through by #400# to allow us to complete the square using integers...

#0 = 400(x^2-4.7x+2.8)#

#color(white)(0) = 400x^2-1880x+1120#

#color(white)(0) = (20x)^2-2(20x)(47)+47^2-1089#

#color(white)(0) = (20x-47)^2-33^2#

#color(white)(0) = ((20x-47)-33)((20x-47)+33)#

#color(white)(0) = (20x-80)(20x-14)#

#color(white)(0) = 40(x-4)(10x-7)#

So:

#x=4" "# or #" "x = 7/10 = 0.7#