Nature of Roots of Quadratic Equation:
The expression, b2 – 4ac, connecting the coefficients of the quadratic equation ax2 + bx + c = 0, with a ≠ 0 determines the nature of the roots α and β of the equation and is known as the discriminant of the equation. It is generally denoted by D.
Several cases may arise as given below-
(i) If D > 0, but not a perfect square, then the roots are real, distinct and irrational.
(ii) If D = k2, k ∈ Q, k ≠ 0, then the roots are real, distinct and rational.
(iii) If D = 0, then the roots are real, rational and equal.
(iv) If D ≥ 0, then the roots are real.
(v) If D < 0, then the roots are imaginary.
(vi) If b = 0, then -b/a = 0 ⇒ α + β = 0, then the roots are equal in magnitudes but opposite in sign.
(vii) If c = a, then c/a = 1 ⇒ αβ = 1, then the roots are reciprocal to each other (provided b ≠ 0 in this case; since b = 0 implies ax2 + a = 0 ⇒ x2 = -1 ⇒ x = ±i).
(viii) If a + b + c = 0, then b = – (a + c)
The equation, then, reduces to ax2 – (a + c) x + c = 0
⇒ ax2 – ax – cx + c = 0 ⇒ ax (x – 1) – c (x – 1) = 0
⇒ (x – 1) (ax – c) = 0 ⇒ x = 1, c/a
Thus, if the sum of the coefficients of a quadratic equation is zero, then one of the roots is unity. In this case, D = b2 – 4ac = {-(a + c)}2 – 4ac = a2 + c2 – 2ac = (a – c)2, which is a perfect square and hence the roots are real distinct and rational.
| Example- Prove that the roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are always real and they are equal if a = b = c. Solution- The given equation is, (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 ⇒ 3x2 – 2(a + b + c)x + (ab + bc + ca) = 0 Its discriminant, D = {-2(a + b + c)}2 – 4 . 3 . (ab + bc + ca) = 4 (a2 + b2 + c2 + 2ab + 2bc + 2 ca) – 12(ab + bc + ca) = 4 (a2 + b2 + c2 – ab – bc -ca) = 2 (2a2 + 2b2 + 2c2 – 2ab -2bc – 2ca) =2{(a – b)2 + (b – c)2 + (c – a)2} Now, for real values of a, b and c, (a – b)2 ≥ 0, (b – c)2 ≥ 0, (c – a)2 ≥ 0 ∴ D ≥ 0. Thus, the roots are real. If now, the roots are equal, then D = 0 ⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0, which is possible only when (a – b)2 = 0, (b – c)2 = 0 and (c – a)2 = 0 ⇒ a = b, b = c and c = a ⇒ a = b = c Hence, for equal roots, a = b = c. |
