3x^2-13x+12=0 plz solve
Answers
Answered by
27
3x²-13x+12 = 0
=> 3x²-9x-4x+12 = 0
=> 3x(x-3)-4(x-3) = 0
=> (3x-4)(x-3) = 0
∴ (3x-4) = 0 or (x-3) = 0
=> x = 4/3 or x = 3
=> 3x²-9x-4x+12 = 0
=> 3x(x-3)-4(x-3) = 0
=> (3x-4)(x-3) = 0
∴ (3x-4) = 0 or (x-3) = 0
=> x = 4/3 or x = 3
Lawliet:
Ahh we both got same answer at same time
I don't think you answering half an hour later is the same time :p
The log makes it obvious man
Answered by
7
3x² - 13x + 12 = 0
3x² - 9x - 4x + 12 = 0
3x(x - 3) -4(x - 3) = 0
(3x - 4)(x - 3) = 0
x = 4/3 0r x = 3
3x² - 9x - 4x + 12 = 0
3x(x - 3) -4(x - 3) = 0
(3x - 4)(x - 3) = 0
x = 4/3 0r x = 3
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