Quadratic Equation(二次方程式的问题)let k be a constant,if A and B are the roots of the equation x^2 - 3x +k = 0,the A^2 + 3B = k 是恒数,如果A和B是x ^2 - 3x+ k= 0的根,A^2 + 3B =
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![Quadratic Equation(二次方程式的问题)let k be a constant,if A and B are the roots of the equation x^2 - 3x +k = 0,the A^2 + 3B = k 是恒数,如果A和B是x ^2 - 3x+ k= 0的根,A^2 + 3B =](/uploads/image/z/14242975-7-5.jpg?t=Quadratic+Equation%28%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B%E5%BC%8F%E7%9A%84%E9%97%AE%E9%A2%98%29let+k+be+a+constant%2Cif+A+and+B+are+the+roots+of+the+equation+x%5E2+-+3x+%2Bk+%3D+0%2Cthe+A%5E2+%2B+3B+%3D+k+%E6%98%AF%E6%81%92%E6%95%B0%2C%E5%A6%82%E6%9E%9CA%E5%92%8CB%E6%98%AFx+%5E2+-+3x%2B+k%3D+0%E7%9A%84%E6%A0%B9%2CA%5E2+%2B+3B+%3D)
Quadratic Equation(二次方程式的问题)let k be a constant,if A and B are the roots of the equation x^2 - 3x +k = 0,the A^2 + 3B = k 是恒数,如果A和B是x ^2 - 3x+ k= 0的根,A^2 + 3B =
Quadratic Equation(二次方程式的问题)
let k be a constant,if A and B are the roots of the equation x^2 - 3x +k = 0,the A^2 + 3B =
k 是恒数,如果A和B是x ^2 - 3x+ k= 0的根,A^2 + 3B =
Quadratic Equation(二次方程式的问题)let k be a constant,if A and B are the roots of the equation x^2 - 3x +k = 0,the A^2 + 3B = k 是恒数,如果A和B是x ^2 - 3x+ k= 0的根,A^2 + 3B =
因为A和B是x ^2 - 3x+ k= 0的两个根,所以A+B=3,即B=3-A
将B带入A^2 + 3B 原式转化为求A^2-3A+9的值
由A是x ^2 - 3x+ k= 0的根可知 A^2-3A= -k
所以原式=9-k