標題:
發問:
Must show the steps very clearly. 1. Given that a and b are intergers, where a is not equal to 0. Prove that the quadratic equation ax^2 +4bx -(a -4b) =0 has rational roots.
最佳解答:
Given a and b are integers. a is not equal to 0. Given a quadratic equation ax^2 + 4bx - (a - 4b) = 0 Delta: (4b)^2 - 4a[-(a - 4b)] = 16b^2 + 4a^2 - 16ab = 4(a^2 - 4ab + 4b^2) = 4(a - 2b)^2 , which is greater or equal to 0. So, the equation has rational roots.
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One A Maths Question發問:
Must show the steps very clearly. 1. Given that a and b are intergers, where a is not equal to 0. Prove that the quadratic equation ax^2 +4bx -(a -4b) =0 has rational roots.
最佳解答:
Given a and b are integers. a is not equal to 0. Given a quadratic equation ax^2 + 4bx - (a - 4b) = 0 Delta: (4b)^2 - 4a[-(a - 4b)] = 16b^2 + 4a^2 - 16ab = 4(a^2 - 4ab + 4b^2) = 4(a - 2b)^2 , which is greater or equal to 0. So, the equation has rational roots.
其他解答:
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