Quadratic equations-1 (XAT-2008)

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),

Answer (E)