Proof: the equation x2 + MX + 1 = 0 of X has two negative real roots if and only if M ≥ 2

Proof: the equation x2 + MX + 1 = 0 of X has two negative real roots if and only if M ≥ 2

The following results are proved: (1) sufficiency: ∵ m ≥ 2, ∵ (?) = M2-4 ≥ 0, the equation x2 + MX + 1 = 0 has real roots. Let two of x2 + MX + 1 = 0 be x1, x2. According to Weida's theorem, ∵ X1 + x2 = - M ≤ - 2, ∵ x1, X2 are both negative roots

A necessary and sufficient condition for the equation MX ^ 2 + (2m + 3) x + 1-m = 0 to have a positive root and a negative root

Let two roots be X1 and X2 respectively. According to Weida's theorem, there are: x1 × x2 = C / A, because X1 and X2 are different signs, so x1 × x2 = C / a ＜ 0. Here, C = 1-m, a = m, so: (1-m) / m ＜ 0. When m ＜ 0, there is 1-m ＞ 0, that is, m ＜ 1. So m ＜ 0 when m ＞ 0, there is 1-m ＜ 0, that is, m ＞ 1

A necessary and sufficient condition for the equation x ^ 2 (2m-1) + m ^ 2 = 0 to be greater than 1 and less than 1

The answer on the first floor is obviously wrong (your range is - 2 & lt; M & lt; 0 & nbsp;, then I can take the special value - 1 to verify. Bar - 1 brings in: X & amp; sup2; = - M & amp; sup2; / (2m-1) = 1 / 3, ∧ x = plus or minus & nbsp; root (3) / 3. Root 3 ≈ 1.732, then divide by 3, obviously less than 1.)
If and only if δ = 0-4m & amp; sup2; (2m-1) > 0 and 2m-1 is less than zero
The solution is m ＜ 0.5. The two roots of the equation can be expressed as - M & amp; sup2; / (2m-1) under x = positive and negative root sign, so that its positive root is greater than 1 and its negative root is less than - 1 (because f (x) = x ^ 2 (2m-1) + m ^ 2 is even function) PS: this radical inequality is difficult to solve and has strong operational skills
If the root sign [- M & amp; sup2; / (2m-1)] is greater than 1, (because both are positive numbers), the square of both sides is: - M & amp; sup2; / (2m-1) > 1. (because there is no limit on the range of M, 2m-1 can not be regarded as a positive number divided by the past), so square again, and get: (m) quartic power / (2m-1) & amp; sup2; > 1. After sorting out and shifting, we get: (m) quartic power - 4m & amp; sup2; + 4m-1 > 0,
Let (m) quartic power - 4m & amp; sup2; + 4m-1 = 0, the quartic equation needs to use the observation method, it can be seen that one root is 1, then the equation becomes: (m-1) (M3 + M & amp; sup2; - 3M + 1) = 0, it can be seen that m3 + M & amp; sup2; - 3M + 1 = 0, one root is 1, it can be decomposed again, it can be found that: (m-1) (m-1) (M & amp; sup2; + 2m-1) = 0, the roots of the equation are m = 1, M = - radical (2) - 1 and & nbsp;
M = root (2) - 1. (next, it's easy to do. You can draw an image according to the rule of odd through and even out. You can also use the Geometer's Sketchpad to find out the value range of m with function image > 0
The scope of this problem m is: (- infinity, - radical (2) - 1) ∪ (radical (2) - 1,1) ∪ (1, positive infinity)
Over, I'm tired to death

What does Matrix I represent in linear algebra?

Identity matrix is a matrix where all diagonal elements are 1 and other elements are 0