root( x2 -3x )= 4x2 - 12x - 3 answer: (3+/- root13)/2 or (6+/- 3root5)/4
√(x2 - 3x) = 4x2 - 12x - 3 = √(x2- 3x) = 4(x2- 3x) - 3
Let (x2- 3x) = A This above equation becomes
√A = 4A - 3 Now squaring both sides we get
A = (4A - 3)2 = A = 16A2 - 24A + 9
= 16A2 - 24A - A + 9 = 0 = 16A2 - 25A + 9 = 0
= 16A2 - 16A - 9A + 9 =0 = 16A(A - 1) - 9(A - 1) =0
= (16A - 9) (A - 1) =0
Thus either 16A - 9 = 0 or A - 1 = 0
CASE 1: when 16A - 9 = 0
= 16(x2- 3x) - 9 = 0 ( A =x2- 3x )
= 16x2 - 48x - 9 = 0 Here a = 16, b = -48, c = -9
Using x = {-b +/- √(b2 - 4ac)}/2a
x = {-(-48) +/- √ (-48)2 - 4x16x-9}/(2x16)
= x = {48 +/-√(2304 + 576)}/32 = x = (48 +/- √2880)/32
= x = (48 +/- 24√5)/32
= x= (6 +/- 3√5)/4
CASE 2: When A - 1 =0
=x2- 3x - 1 = 0. ( A =x2- 3x)
Again usingx = {-b +/- √(b2- 4ac)}/2a
x = {-(-3) +/-√(-3)2 - 4x1x-1}/(2x1)
= x = {3 +/-√(9 + 4)}/2
= x = (3+/- √13)/2