Solution set of inequality 2x & sup2; - 11x + 12 < 0

Solution set of inequality 2x & sup2; - 11x + 12 < 0

(2x-3)(x-4)
3/2 < x < 4
Solving inequality 2x & sup2; - 11x + 5 > 0
2x² -11x+5＞0
(2x-1)(x-5)>0
Opening up
X5
X2.5
Original formula = (2x-1) * (X-5) > 0
=> X>5 or X< 1/2
1.2x & # 178; - 2x-3 = 0 (solved by formula method) 2.4 (X-2) & # 178; = 9 (2x-1) & # 178;
1. 2X & # 178; - 2x-3 = 0 (solved by formula) 2 (x-1 / 2) & # 178; = 7 / 2, (x-1 / 2) & # 178; = 7 / 4, so X-1 / 2 = ± √ 7 / 2, that is, x = (1 ± √ 7) / 2.2, 4 (X-2) & # 178; = 9 (2x-1) & # 178;, 2 (X-2) = ± 3 (2x-1), 2X-4 = ± (6x-3), x = - 1 / 4, or x = 7 / 8
If the two square roots of a positive number a are 2m-3 and 5-m respectively, find the value of A
The two square roots of ∵ a positive number a are 2m-3 and 5-m respectively, and ∵ 2m-3 + 5-m = 0. The solution is m = - 2, then 2m-3 = - 7, and a = 49
It is known that the quadratic equation KX & # 178; - 2 (K + 1) x + K-2 has two unequal real roots, and the range of K is obtained
According to the discriminant method, when the quadratic equation of one variable has unequal real roots:
△ = b²-4ac
= 4(k+1)² - 4k(k-2)
= 4k²+8k+4-4k²+8k
=16k+4 > 0
Therefore:
k > -1/4
Solution equation: 10.5 + X + 21 = 56
x=56-21-10.5
x=56-31.5
x=24.5
x=24.5。。。。。。。。。。。。。。。。。
Given that the square roots of 4m + 1 are 5 and - 5, and the square roots of 5m + N + 1 are 6 and - 6, find the value of 2m-3n
It is known that the square roots of 4m + 1 are 5 and - 5
So 4m + 1 = 25 m = 6
The square roots of 5m + N + 1 are 6 and - 6
So 5m + N + 1 = 36 = 31 + n
So n = 5
So 2m-3n = 12-15 = - 3
4m+1=5^2=25 m=6
5m+n+1=6^2=36 n=5
2m-3n=2*6-3*5=12-15=-3
1. Proof: for any real number m, there are two unequal real number roots of the equation x square - (m-1) x-3 (M + 3) = 0
△=b^2-4ac
=〔-（m-1)〕^2-4*〔-3(m+3)〕
=m^2-2m+1+12m+36
=m^2+10m+37
=(m+5)^2+12
No matter what the value of M is, there is (M + 5) ^ 2 ≥ 0,
So △ = (M + 5) ^ 2 + 12 ＞ 0
Therefore, for any real number m, the equation has two unequal real roots
∵△=（m-1）^2+4x3（m+3）
=m^2+10m+37
=（m+5）^2+12＞0
The equation of X with respect to X has two unequal real roots: Square - (m-1) x-3 (M + 3) = 0.
34: x = 56 2.5: x = 4
34:x=56
x=34÷56
x=17/28
2.5：x=4
x=2.5÷4
x=0.625
If the equation x ^ 2 + 2x + 2m = 0 has two real roots with the same sign, then the value range of M is?
Urgent need
According to Weida's theorem
We get a + B = - 2
We can see that a and B are both negative numbers
2m = a * B should be positive, so m is greater than 0
According to the discriminant b2-4ac of roots
The solution m is less than or equal to 1 / 2 (the original formula has real roots)
in summary
M greater than 0 and less than or equal to 1 / 2
There should be another one upstairs
Because he didn't say the two were not equal
First of all, we need to satisfy △ > 0, so M0 is enough
That is 2m > 0
To sum up, 0