Question

The roots of the equation
(p - 2)x2 + 2(p - 2)x + 2 = 0 are not real when-
(1) pe [1,2] (2) p e[2, 3]
(3) pe (2,4) (4) Pe[3, 4]​

Answer 1

Answer:

(3).

Step-by-step explanation:

roots are not real, when discriminant of an equation is less than zero.

here,

$d = {b}^{2} - 4ac \\ d = (2(p - 2)) {}^{2} - 4(2)(p - 2) < 0 \\ 4(p - 2) {}^{2} < 8(p - 2) \\ (p - 2) {}^{2} - 2(p - 2) < 0 \\ (p - 2)(p - 2 - 2) < 0 \\ (p - 2)(p - 4) < 0$

therefore,

p€ (2,4)