## Nature of Roots of Quadratic Equation:

The expression, b^{2} – 4ac, connecting the coefficients of the quadratic equation ax^{2} + 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 = k^{2}, 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 ax^{2} + a = 0 ⇒ x^{2} = -1 ⇒ x = ±*i*).

(viii) If a + b + c = 0, then b = – (a + c)

The equation, then, reduces to ax^{2} – (a + c) x + c = 0

⇒ ax^{2} – 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 = b^{2} – 4ac = {-(a + c)}^{2} – 4ac = a^{2} + c^{2} – 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 ⇒ 3x ^{2} – 2(a + b + c)x + (ab + bc + ca) = 0Its discriminant, D = {-2(a + b + c)} ^{2} – 4 . 3 . (ab + bc + ca)= 4 (a ^{2} + b^{2} + c^{2} + 2ab + 2bc + 2 ca) – 12(ab + bc + ca)= 4 (a ^{2} + b^{2} + c^{2} – ab – bc -ca)= 2 (2a ^{2} + 2b^{2} + 2c^{2} – 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. |

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