If 0 < p <
1, then the roots of the equation (1-p)x

^{2}+4x+p = 0 are
(A)Both zero

(B) Imaginary

(C)Real and
both positive

(D)Real and of
opposite sign

(E) Real and
both negative

Solution follows here:

__Solution:____Funda:__

Roots of the quadratic equation ax

^{2}+bx+c=0 are -b±√(b^{2}-4ac) / 2a
If b

^{2}-4ac = 0, both roots are real and equal,
If b

^{2}-4ac > 0, both roots are real and different,
If b

^{2}-4ac < 0, both roots are imaginary.
First
let us find b

^{2}-4ac = 4^{2}-4(1-p)(p) = 16-4p+4p^{2}= 4(p^{2}-p+4)
0
< p < 1 => Both p

^{2},p are fractions=> p^{2}-p is a also fraction=> p^{2}-p+4 > 0
=>
Both the roots are real ---(1)

Roots
= -4±√[4(p

^{2}-p+4)] / 2(1-p)
As
0 < p < 1, p

^{2 }< p => p^{2}-p < 0 => p^{2}-p+4 < 4 => 4(p^{2}-p+4) < 16 => √[4(p^{2}-p+4)] < 4
Observe
this: -4±√[4(p

^{2}-p+4)]
As
the second part (the part after ‘±’) is positive and less than 4, -4+√[4(p

^{2}-p+4)] is negative and -4-√[4(p^{2}-p+4)] is also negative.
=>
Both the roots are negative ---(2)

From
(1) and (2),

**Answer (E)**
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