When the real number m takes what value, the equation 4x + (m-2) x + (m-5) = 0 has: (1) the absolute value of positive root is greater than that of negative root When m is a real number, the equations 4x + (m-2) x + (m-5) = 0 are as follows: (1) Is the absolute value of positive root greater than that of negative root? (2) Both are greater than 1?

When the real number m takes what value, the equation 4x + (m-2) x + (m-5) = 0 has: (1) the absolute value of positive root is greater than that of negative root When m is a real number, the equations 4x + (m-2) x + (m-5) = 0 are as follows: (1) Is the absolute value of positive root greater than that of negative root? (2) Both are greater than 1?

4x^2+(m-2)x+(m-5)=0
Let the two parts of the equation be X1 and X2 respectively
(1)
The equation has two real roots
∴△=(m-2)^2-4*4(m-5)=m^2-20m+84>0
The solution is M14
When M14, the equation has two real roots
(2)
The equation has a positive root and a negative root
∴△=(m-2)^2-4*4(m-5)=m^2-20m+84>0
x1x2=(m-5)/4