Question

Q) Let b be
non- zero real number. Suppose the
quadratic equation 2x^2 + bx + 1/b = 0 has two distict
real rools. Then,
(A)b+1/b>5/2

(B) b+1/b<5/2

(C) b^2+1/b^2<4

(D)b^2-3b>-2

In the ans key it is given the answer as (d)
Please explain how
we get the answer.​

Answer 1

Step-by-step explanation:

The given equation is 2x2+bx+b1=0, comparing it with Ax2+Bx+C=0. We get

A=2,B=b and C=b1

⇒  B2−4AC=(b)2−4(2)×b1

⇒  B2−4AC=b2−b8

Now,

Discriminant>0                       [ For real roots ]

b2−b8>0                 [ For real roots ]

bb3−(2)3>0

b(b−2)(b2+2b+4)>0  

b2+2b+4 always positive

bb−2>0

b−2>0,b>0                       and        b−2<0,b<0