## Monday, 7 November 2011

If 0 < p < 1, then the roots of the equation (1-p)x2+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 ax2+bx+c=0 are -b±√(b2-4ac) / 2a
If b2-4ac = 0, both roots are real and equal,
If b2-4ac > 0, both roots are real and different,
If b2-4ac < 0, both roots are imaginary.
First let us find b2-4ac = 42-4(1-p)(p) = 16-4p+4p2 = 4(p2-p+4)
0 < p < 1 => Both p2,p are fractions=> p2-p is a also fraction=> p2-p+4 > 0
=> Both the roots are real                                                              ---(1)
Roots = -4±√[4(p2-p+4)] / 2(1-p)
As 0 < p < 1, p2 < p => p2-p < 0 => p2-p+4 < 4 => 4(p2-p+4) < 16 => √[4(p2-p+4)] < 4
Observe this: -4±√[4(p2-p+4)]
As the second part (the part after ‘±’) is positive and less than 4, -4+√[4(p2-p+4)] is negative and -4-√[4(p2-p+4)] is also negative.
=> Both the roots are negative                                                                   ---(2)
From (1) and (2),