If (x²-5x+6)² + (x²-7x+12)² =0 the the value of x is?
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Answer:
Step-by-step explanation:
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Answer:
x = 3, 3 + i, 3 - i
Step-by-step explanation:
(x² - 5x + 6)² + (x² - 7x+12)² = 0
=> { (x - 3) (x -2) }² + { (x - 3) (x - 4) }² = 0
=> (x - 3)² {(x - 2)² + (x - 4)² } = 0
=> (x - 3)² (x² - 4x + 4 + x² - 8x +16) = 0
=> (x - 3)² (2x² - 12x + 20) = 0
=> 2(x - 3)² (x² - 6x + 10) = 0
=> (x - 3)² (x² - 6x + 10) = 0
=> (x - 3)² {(x² - 6x + 9) + 1} = 0
=> (x - 3)² {(x - 3)² + 1} = 0
From above equation:
i) (x - 3)² = 0
=> x - 3 = 0
=> x = 3
ii) {(x - 3)² + 1} = 0
=> (x - 3)² = -(1)²
=> x - 3 = ± i
=> x = 3 ± i
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