Question

|x²-4x+3| = -|x²-5x+4|

Topic - Modulus Equations

Do it without solving it

D@mn easy -,- ​

Answer 1

Given :-

| x² - 4x + 3 | = - | x² - 5x + 4 |

To Find :-

The solution of the given without solving it.

Solution :-

As the LHS Is equal to RHS . So ,

Let , | x² - 4x + 3 | = | x² - 5x + 4 | = y ----- ( i )

Now. , as per the given situation ,

| x² - 4x + 3 | = - | x² - 5x + 4 |

=> y = - y [ By ( i ) ]

=> But , y ≠ - y . So , the given situation is not possible .

Note that , 0 = -0 . But , this situation is not possible on x = 0 also ! By checking on x at 0 :-

| x² - 4x + 3 | = - | x² - 5x + 4 |

=> Put , x = 0

=> | 0² - 4 × 0 + 3 | = - | 0² - 5 × 0 + 4 |

=> | 3 | ≠ - | 4 |

So , Verified that this situation is not possible at x = 0 .

Edit * :-

As you can see that this situation is only possible if and only | x² - 4x + 3 | = - | x² - 5x + 4 | = 0 = -0 . We didn't get | x² - 4x + 3 | = - | x² - 5x + 4 | = 0 = -0 by putting x = 0 . So we have to find such value / values of x such that | x² - 4x + 3 | = - | x² - 5x + 4 | = 0 = -0 . Because the only real number with positive sign is equals to itself with negative sign is 0 .

I hope that you will understand what I want to explain .

So , This is only possible when x = 1.

So , Let's put x = 1 ,

| x² - 4x + 3 | = - | x² - 5x + 4 |

=> | 1² - 4 × 1 + 3 | = - | 1² - 5 × 1 + 4 |

=> | -4 + 4 | = - | -5 + 5 |

=> | 0 | = - | 0 |

=> 0 = -0

Henceforth , Proved !

Answer 2

Answer:

Your answer is x=1 and x=7/2.

Hope this helps ü mate ✔️

Step-by-step explanation:

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