Hello together, a possible alternative put a 2 out of every bracket and substitute y=x/2. Now you have y(y+1)(y+2)(y+3)=9/16. The point of it is that we know add a 1 on both sides, both sides become pure squares: (x^2+3x+1)^2=25/16. After taking the square root on both sides one is left with two quadratic equations.
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This was a fourth-grade equation. The reason why we have three solutions is because the one x=-3 is a double solution.
It has become very easy after your explanation
Hello together, a possible alternative put a 2 out of every bracket and substitute y=x/2. Now you have y(y+1)(y+2)(y+3)=9/16. The point of it is that we know add a 1 on both sides, both sides become pure squares: (x^2+3x+1)^2=25/16. After taking the square root on both sides one is left with two quadratic equations.
Nice explaination
Very good explanation mam 👏 👌
good
А почему нельзя найти корни квадратного уравнения напрямую по теореме Виета, например?
Этого так не решится. Неправильно решения