If m is positive but not greater than 3 then show that the roots of the equation (m-2)x² - (8-2m)x - (8-3m) = 0 are real . find the range of values of m for which one root is positive and the other root is negative

Dear Student,

(m-2) x² - (8-2m)x - (8-3m) = 0

Comparing it with ax² +bx +c = 0

a = (m-2) , b = -(8-2m) , c = -(8-3m)

D = b² - 4ac
For roots to be real D must be greater than equal to 0.
Therefore,

D = (8-2m)² - 4(m-2) (-(8-3m))
   = 64 - 32m + 4m² + 4 (8m - 3m² - 16 + 6m^​2 ) 
  = 4 (16 - 8m + m² + 8m + 3m² - 16)
 = 4 (4m² ) 
= 16 m²  
Hence, D is grater than 0 and the roots are real 

Now, one root is positive and other negative
Hence, product of roots is less than 0

Product of roots = $\frac{\mathrm{c}}{\mathrm{a}} = \frac{-(8-3\mathrm{m})}{(\mathrm{m}-2)}$ < 0
$\frac{(3\mathrm{m}-8)}{(\mathrm{m}-2)} <0$

$\Rightarrow 2<\mathrm{m}<\frac{8}{3}$

Regards