It is known that the integer m satisfies 6 ＜ m ＜ 20. If the quadratic equation MX2 - (2m-1) x + m-2 = 0 with respect to X has a rational root, the value of M and the root of the equation are obtained

It is known that the integer m satisfies 6 ＜ m ＜ 20. If the quadratic equation MX2 - (2m-1) x + m-2 = 0 with respect to X has a rational root, the value of M and the root of the equation are obtained

According to the meaning of the question, m ≠ 0, if the equation has rational roots, then △ is a complete square number. ∵ (?) = (2m-1) 2-4m × (m-2) = 4m + 1, and ∵ integer m satisfies 6 ＜ m ＜ 20, ∵ 4m + 1 = 49, that is, M = 12. Then the original equation becomes: 12x2-23x + 10 = 0, ∵ x = 23 ± 492 × 12 = 23 ± 724, ∵ X1 = 23, X2 = 54

(m-1) x ^ 2 + X + M-1

∵ (m-1) x & # 178; + X + (m-1) = 0 is a quadratic equation with one variable
The quadratic coefficient is not 0
That is, M-1 ≠ 0
∴m≠1
∴m∈(-∞,1)∪(1,+∞)

Verification: no matter what real number x takes, the value of polynomial (x-1) * (x-3) * (x-4) * (X-6) is not less than - 9

(x-1) * (x-3) * (x-4) * (X-6) = [(x-1) (X-6)] [(x-3) (x-4)] = [(x ^ 2-7x) + 6] [(x ^ 2-7x) + 12] = (x ^ 2-7x) ^ 2 + 18 (x ^ 2-7x) + 72 let t = x ^ 2-7x, then (x-1) * (x-3) * (x-4) * (X-6) = T ^ 2 + 18t + 72 = (T + 9) ^ 2-9 > = - 9

|Inequality solution of 3 + X / 2x-1 | > = 1

(3+x)/(2x-1)=1
(3x+2)/(2x-1)

Given that a, B and C are trilateral lengths of △ ABC, this paper proves that the equation CX & # 178; - (a + b) x + C / 4 = 0 has two unequal real roots
Because C is not equal to 0, the original equation is a quadratic equation with one variable,
Its discriminant △ = (a + b) ^ 2-4c (C / 4) = (a + b) ^ 2-C ^ 2 = (a + B + C) (a + B-C)
Because a, B and C are the three sides of △ ABC, a + B + C > 0, a + B-C > 0
So the discriminant △ > 0
So the original equation has two unequal real roots
Are there two unequal positive roots or
Have two unequal negative roots?

Let two be X1 and X2 respectively
x1+x2=(a+b)/c
x1x2=1/4
If the product of two is greater than 0, the two have the same sign and are not equal to 0;
If the sum of the two roots is greater than 0, both of them are positive roots (if they are both negative roots, then the sum of the two roots is negative)

Finding the solution set of inequality (2x-3) (x + 1) > 0
Don't tell me quadratic function image I haven't learned!

Multiplication greater than 0
Then all are greater than 0 or less than 0
2x-3>0,x+1>0
Then x > 3 / 2, x > - 1
Take the big from the big
So x > 3 / 2
2x-3

Let a, B and C be the three sides of △ ABC, then the square of equation CX + (a + b) x + 4 / 4 C = 0 has two unequal negative real roots

△=（a+b)^2-c^2=(a+b+c)(a+b-c)
The sum of any two sides of the triangle is greater than the third side, that is, a + B-C > 0
So we have △ > 0
The equation has two unequal real roots
Another two products = 1 / 4, so the two have the same sign
Two sum = - (a + b) / C

If the solution set of inequality 13 (2x-k) > x-k is x < - 23, then k = 1___ ．

If we remove the brackets, we get 23x-k3 ＞ x-k; if we shift the term, we get x ＜ 2K; ∵ its solution set is x ＜ - 23, ∵ 2K = - 23, ∵ k = - 13