How do you solve by completing the square #3x^2-2x-3=0#? and solve by quadratic formula 2x^2-4x+1=0 find the length and width of a rectangle whose perimeter is 16 units and whose area is 11 square units?

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Answer 1 · 2015-04-03T19:54:01

No. 1
#3x^2 - 2x - 3 = 0#

#3(x^2 - 2/3x -1) = 0#

Now take the coefficient of #x# and divide it by #2# and square it,
#-2/3 -> -1/3 -> +1/9#

then add it and subtract it,

#3(x^2 - 2/3x + 1/9 -1/9 - 1) = 0#

Since #x^2 -2/3x + 1/9 = (x - 1/3)^2#

#=> 3[(x - 1/3)^2 -1/9 -1] = 0#

#=> (x - 1/3)^2 - 10/9 = 0#

#=> (x - 1/3 - sqrt(10/9))(x - 1/3 + sqrt(10/9)) = 0#

#=> x - 1/3 - sqrt(10/9) = 0 => x = (1 + sqrt(10))/3#

#=> x - 1/3 + sqrt(10/9) = 0 => x = (1 - sqrt(10))/3#

No. 2
#2x^2 - 4x + 1 = 0#

#x = (-(-4) +- sqrt((-4)^2 -4*2*1))/(2*2)#

#=> x = (4 +- sqrt(8))/4 = (4 +- 2sqrt(2))/4 = 2 +-sqrt(2)/2#

No. 3
Perimeter : #2(L + W) = 16# ... (1)

Area : # LxxW = 11# ... (2)

Solve the two equations simultaneously,

In (1), #L + W = 8 => L = 8 - W#

sub #L# in (2)

#=> (8 - W)*W = 11 => W^2 -8W + 11 = 0#

#=> W = (8 +- sqrt(64 - 44))/2 => W = 4 +- sqrt(5)#

Back to equation (1), put #W# in (1),

#=> L = 8 - W => L = 8 - 4 +-sqrt(5) => L = 4 +- sqrt(5)#

Note : when #W = 4 - sqrt(5)# , #L = 4 + sqrt(5)#