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

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