Question

| x^2 – 4x + 3 | = 3 | x^2 – 5x + 4 |

Answer 1

Answer ;

|x² - 4x + 3| = 3|x² - 5x + 4|

Opening the modulus

x² - 4x + 3 = 3x² - 15x + 12

-2x² + 11x - 9 = 0

2x² - 11x + 9 = 0

2x² - (2 + 9)x + 9= 0

2x² - 2x - 9x + 9= 0

2x(x - 1) -9 (x - 1) = 0

(2x - 9)(x - 1) = 0

x = -9/2 or x = 1

Hence the value of x = -9/2 or 1

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Answer 2

Given :   | x^2 – 4x + 3 | = 3 | x^2 – 5x + 4 |  

To find : x

Solution:

| x² – 4x + 3 | = 3 | x²– 5x + 4 |

| f(x) |  = f(x)  if  f(x)  ≥ 0  &  - f(x)  if f(x) < 0  

x² – 4x + 3  = (x - 3)(x - 1)          for   x ≥ 3  &

| x² – 4x + 3 | = x² – 4x + 3  for   x ≥ 3  &  x ≤ 1

| x² – 4x + 3 | = -(x² – 4x + 3) for   1  < x  < 3

x²– 5x + 4 = (x - 4)(x - 1)

| x²– 5x + 4 | = x²– 5x + 4   x ≥ 4  &  x ≤ 1

| x²– 5x + 4 | = -(x²– 5x + 4)   1  < x  <  4

Both + ve & - ve together  

for  x ≥ 4  &   x ≤ 1  ( here both are + ve)  ,     1  < x  < 3 ( here both are -ve)

x² – 4x + 3  = 3( x²– 5x + 4)

=> 2x² - 11x + 9   = 0

=> 2x² - 2x -9x +  9   = 0

=> 2x(x - 1) - 9(x - 1) = 0

=> (2x - 9)(x - 1)  = 0

=> x = 9/2  , x  = 1

both are in between x ≥ 4  &   x ≤ 1   , 1  < x  < 3  

for   3  ≤ x  < 4    ( one is + ve & another is - ve )

x² – 4x + 3  = -3( x²– 5x + 4)

=> 4x² - 19x + 15 = 0

=> 4x² - 4x - 15x + 15 = 0

=> 4x(x - 1) -15(x - 1) = 0

=> (4x - 15)(x  -1) = 0

=> x = 15/4  ,  1

x = 15/4   is in between   3  ≤ x  < 4

x = 1  , 15/4  , 9/2

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