If the solution set of inequality ax2-5x + b > 0 is (- 3, - 2), find the solution set of inequality bx2-5x + a > 0 Urgent!

If the solution set of inequality ax2-5x + b > 0 is (- 3, - 2), find the solution set of inequality bx2-5x + a > 0 Urgent!

Obviously, A0 is
6x2+5x+1

Solve the inequality x square - 4x + 3 > = 0

(x-1)(x-3)>=0,
x>=3 or x

Given the function f (x) = {3 ^ x, 0 ≤ x ≤ 1; X ^ 2-4x + 4, x > 1, then inequality 1

(0,1) ,(3,4)

Solve the following equation by factorization
First question, 3x & # 178; - 12x = - 12
Question 2, 4x & # 178; - 144 = 0
Question 3, 3x (x-1) - 2 (x-1)
Question 4, (2x-1) &# 178; - (3-x) &# 178;

5x (x-3) = 6-2x5x (x-3) = 2 (3-x) 5x (x-3) + 2 (x-3) = 0 (x-3) (5x + 2) = 0x-3 = 0 or 5x + 2 = 0x1 = 3; x2 = - 2 / 52.9 (X-2) &# 178; = 4x & # 178; [3 (X-2)] &# 178; - (2x) &# 178; = 0 [3 (X-2) + 2x] [3 (X-2) - 2x] = 0 (5x-6) (X-6) = 0 or X-6 = 0x1 = 6 / 5

Solve the equation about X by factorization
x^2-bx-2（b^2）=0

The following equation is solved by factorization; t (T-4 √ 3) = - 12

T(T-4√3)=-12
T^2-4√3T+12=0
(T-2√3)^2=0
T=2√3

What is the interval of the root of equation x3-3x-3 = 0
The answer is (2,3)

Let f (x) = x3-3x-3, then the function f (x) increases monotonically, bringing x = 2, x = 3 into f (2) 0, so the root is in the interval (2,3)

The root of equation x3-2x2 + 3x-6 = 0 on interval [- 2,4] must belong to interval
A.[-2,1]
B.[2.5,4]
C.[1,7/4]
D.[7/4,2.5]
There are so many options, I want to know how to do it!

Is X3 the third power of X?
Choose D, my exclusion method
Let y = x3-2x2 + 3x-6
First, the derivative is y '= 3x2-4x + 3, from b2-4ac > 0, the derivative is always greater than 0, so the function increases
So it is also constant increasing on [- 2,4]. If we take x = 1, we get y = - 40, so we exclude ABC and choose D

(there will be high delivery after finishing)
Numerical calculation method: the dichotomy method is used to find the root of 3x & # 179; - X & # 178; - 1 = 0 in the interval [0,1], and the required accuracy is 0.005

f(x)=3X³-X²-1
Step 1:
f(0)=3*0³-0²-1=-1＜0
f(1)=3*1³-1²-1=2＞0
Step 2:
F (0.5) = 3 * 0.5 & # 179; - 0.5 & # 178; - 1 = 0.375-0.25-1 = - 0.875 ＜ 0, we can see that there is no big difference between X = 0.5 and f (0)
f(0.9)=3*0.9³-0.9²-1=2.187-0.81-1=0.377＞0
Step 3:
F (0.85) = 3 * 0.85 & # 179; - 0.85 & # 178; - 1 = 1.842375-0.7225-1 = 0.119875 ＞ 0
F (0.8) = 3 * 0.8 & # 179; - 0.8 & # 178; - 1 = 1.536-0.64-1 = - 0.104 ＜ 0, yes
Step 4:
f(0.83)=3*0.83³-0.83²-1=1.715361-0.6889-1=0.026461＞0
f(0.82)=3*0.82³-0.82²-1=1.654104-0.6724-1=-0.018296＜0
Step 5:
f(0.825)=3*0.825³-0.825²-1=1.684546875-0.680625-1=0.003921875＞0
f(0.824)=3*0.824³-0.824²-1=1.678428672-0.678976-1=-0.000547328＜0
Enough, the root of 3x & # 179; - X & # 178; - 1 = 0 in the interval [0,1] is 0.824 ＜ x ＜ 0.825
In fact, x = 0.82412

Solution equation: 8x + 5 * (x + 8.5) = 88

8x+5*(x+8.5)=88
8x+5x+42.5=88
13x=88-42.5
13x=45.5
x=45.5/13
X=3.5