# The equation 2x2 + 2(p + 1)x + p = 0, where p is real, always has roots that are *  Equal  *  Equal in magnitude but opposite in sign  *  Irrational  *  Real  *  Complex Conjugates

vijay narang
27 Points
14 years ago

The value of the discriminant of a quadratic equation will determine the nature of the roots of a quadratic equation.

The discriminant of a quadratic equation ax2 + bx + c = 0 is given by b2 - 4ac.

* If the value of the discriminant is positive, i.e. greater than '0', then the roots of the quadratic equation will be real.
* If the value of the discriminant is '0', then the roots of the quadratic equation will be real and equal.
* If the value of the discriminant is negative, i.e. lesser than '0', then the roots of the quadratic equation will be imaginary. The two roots will be complex conjugates of the form p + iq and p - iq.

Using this basic information, we can solve this problem as shown below.

In this question, a = 2, b = 2(p + 1) and c = p

Therefore, the disciminant will be (2(p + 1))2 - 4*2*p = 4(p + 1)2 - 8p
= 4[(p + 1)2 - 2p]
= 4[(p2 + 2p + 1) - 2p]
= 4(p2 + 1)

For any real value of p, 4(p2 + 1) will always be positive as p2 cannot be negative for real p.

Hence, the discriminant b2 - 4ac will always be positive.

When the discriminant is greater than '0' or is positive, the roots of a quadratic equation will be real.