If roots if the equation (m-2)x2-(8-2m)x-(8-3m) =0 are opposite in sign then number of intergral values of m is /are
Answers
Concept:
The sum of a quadratic equation's roots is -b/a, where b is the coefficient of the variable with degree 1, and a is the coefficient of the variable with degree 2, and the product of zeroes is c/a, where c is the constant term, and a is the coefficient of the variable with degree 2.
Given:
Given quadratic equation is (m-2)x² - (8-2m)x - (8-3m) = 0
Roots of the quadratic equation are opposite in sign.
Find:
We have to find number of integral values of m.
Solution:
Assume that t and -t are the equation's roots.
If we do sum of roots
t + (-t) = -b/a
t - t = (8-2m)/(m-2)
0 = (8-2m)/(m-2)
0 = 8 - 2m
8 = 2m
m = 4
from here m = 4 but m ≠ 2 (as denominator can not be 0)
If we do products of roots
t × (-t) = c/a
-t² = - (8-3m) / (m-2)
t² = (8-3m) / (m-2)
t = √( (8-3m) / (m-2) )
means,
(8-3m) / (m-2) ≥ 0
By applying wavy curve method we find out that
2 > m ≥ 8/3
2 > m ≥ 2.66
Hence we can say that m ≠ 4 as well.
Moreover, as there is no integral value lie between 2 and 2.66
Hence, there will be 0 integral values of m.
#SPJ3