How do you solve #2( 4- x ) ^ { 2} + 3= 5#?

2 Answers
LM
Jan 28, 2018

#x = 3, x = 5#

Explanation:

#2(4-x)^2+3=5#

#(4-x)^2 = (4-x)(4-x)#

use the FOIL method to expand brackets:

#(4-x)(4-x) = (4*4) + (4 * -x) + (-x * 4) + (-x * -x)#

#=16 -4x -4x + x^2#

#= x^2 - 8x + 16#

set the RHS of the equation to #0:#

#2(4-x)^2 = 2(x^2-8x+16) = 2x^2-16x+32#

#2(4-x)^2+3=5#

#2x^2 - 16x + 32 + 3 = 5#

#2x^2 - 16x + 35 = 5#

#2x^2 - 16x + 30 = 0#

then use the quadratic formula:

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

#ax^2 + bx + c = 2x^2 - 16x + 30#

#a = 2, b = -16, c = 30#

#x = (16 +- sqrt(256-(4*2*30)))/(4)#

#x = (16 +- sqrt(16))/(4)#

#x = (16 +- 4)/4#

#x = 12/4, x = 20/4#

#x = 3, x = 5#

Rhys
Jan 28, 2018

Alternate method....

Explanation:

The other two answers use expansions, and then solving, but this method is slightly different but give the same solutions...

#2(4-x)^2 +3 = 5 #

Subtract 3 from both sides...

#2(4-x)^2 = 2 #

Divide by 2 on each side...

#(4-x)^2 = 1 #

Square root each side:

#sqrt( (4-x)^2 ) = sqrt(1) #

#=> 4-x = pm 1 #

Rearanaging....

#=> -x = -4 pm 1 #

#=> x = - ( -4 pm 1 )#

#=> x = 4 pm 1 #

#x = { 4-1 , 4+1 } = color(red)( underline({ 3 ,5 }) #

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