Given that x 1 and x 2 are two positive integer roots of the equation x 2-m x + 5 (m-5) = 0, and x 1 + x 2 = 7, find the value of M

Given that x 1 and x 2 are two positive integer roots of the equation x 2-m x + 5 (m-5) = 0, and x 1 + x 2 = 7, find the value of M

According to the meaning of the question, we can get △ = M2-4 × 5 (m-5) = m2-20m + 100 = (M-10) 2 ≥ 0, x = m ± (m − 10) 2, the solution is x = 5 or x = m-5, ∵ x1, X2 are the two positive integer roots of the equation x2 MX + 5 (m-5) = 0, ∵ m-5 > 0, the solution is m > 5, when X1 = 5, X2 = m-5, 10 + m-5 = 7, the solution is m = 2 (rounding off), when X1 = m-5, X2 = 5, 2 (m-5) + 5 = 7, the solution is m = 6, ∵ m is 6

Do it with Veda's theorem: x ^ 2 + 2x + M-1 = 0, what is the value of M, and the equation has two unequal real roots?

Let two be a and B
Then we know from the Veda theorem that:
a+b=-2,ab=m-1
So (a-b) ^ 2 = (a + b) ^ 2-4ab = 4-4 (m-1) = 8-4m
Because a and B are not equal, 8-4m > 0
That is: M

If the equation AX2 + BX + C = 0 (a ≠ 0) has two real roots x1, X2, then there are & nbsp; X1 + x2 = − Ba, x1 · x2 = ca. this theorem is called Veda's theorem. According to Veda's theorem, the following problems can be solved: given that LGM and LGN are the two real roots of the equation 2x2-4x + 1 = 0, then (1) find the value of Mn; (2) find the value of lognm + logmn

(1) Let LGM and LGN be the two real roots of the equation 2x2-4x + 1 = 0, then we can get LGM + LGN = 2 and LGM · LGN = 12 from WIDA's theorem. Therefore, LG (MN) = 2, Mn = 100. (2) since lognm + logmn = lgmlgn + lgnlgm = (LGM) 2 + (LGN) 2lgm · LGN = (LGM + LGN) 2 − 2lgm · lgnlgm ·