求高手解答--數學題

Let the solutions of :x² - ax + b > 0 is x >3 or x < m ,x² + bx - 4 ≤ 0 is n ≤ x ≤ 4 ,then
x² - ax + b > 0
becomes
(x - m) (x - 3) > 0
x² - (m+3)x + 3m > 0Comparing coefficients :
m+3 = a
{
3m = b ;

x² + bx - 4 ≤ 0
becomes
(x - 4) (x - n) ≤ 0
x² - (n+4)x + 4n ≤ 0Comparing coefficients :
- (n+4) = b
{
4n = - 4
So
n = - 1 and b = - 3

then
m = - 1 , a = 2
Verify :Slove
x² - 2x - 3 > 0 ==> (x - 3)(x + 1) > 0 ==> x > 3 or x ≤ - 1
{
x² - 3x - 4 ≤ 0 ==> (x - 4)(x + 1) ≤ 0 ==> - 1 ≤ x ≤ 4==>3 < x ≤ 4