It is proved that the quadratic equation x 2 + (a + 1) x + 2 (A-2) = 0 must have two unequal real roots

It is proved that the quadratic equation x 2 + (a + 1) x + 2 (A-2) = 0 must have two unequal real roots

It is proved that: ∵ = (a + 1) 2-4 × 2 (A-2) = a2-6a + 17 = (A-3) 2 + 8, ∵ (A-3) 2 ≥ 0, ∵ (A-3) 2 + 8 > 0, that is to say, ∵ the original equation must have two unequal real roots

It is known that the quadratic equation X2 - (K-3) x-k2 = 0 (1) with respect to X has two unequal real roots no matter what the value of K is
The quadratic equation X2 - (K-3) x-k2 = 0 with respect to X is known
(1) It is proved that no matter what value k is, the original equation always has two unequal real roots
(2) If X1 and X2 are two of the original equations, and the absolute value of x1-x2 = 2 times the root sign 2, find the value of K

(1) Therefore, no matter what the value of K is, the original equation always has two unequal real roots
(2) | x1-x2 | ^ 2 = 8, (x1 + x2) ^ 2-4 * X1 * x2 = 8, because X1 + x2 = K-3, X1 * x2 = - K ^ 2
So the solution is k = 1 or K = 1 / 5

The interval where the equation X-1 = lgx must have a root is ()
A. （0.1，0.2）B. （0.2，0.3）C. （0.3，0.4）D. （0.4，0.5）

Let f (x) = x-1-lgx, then f (0.1) = 0.1-1-lg0.1 = 0.1 ＞ 0, f (0.2) = 0.2-1-lg0.2 = 0.2-1 - (lg2-1) = 0.2-lg2, ∵ lg20.2 = LG2 & nbsp; 10.2 = LG32 ＞ LG10 = 1; ∵ LG2 ＞ 0.2; f (0.2) ＜ 0; similarly: (0.3) = 0.3-1-lg0.3 - (lg3-1) = 0