## Properties of Roots of a Quadratic Equation:

(1) *A quadratic equation cannot have more than two roots.*

**Proof:** If possible, let the quadratic equation ax^{2} + bx + c = 0, a ≠ 0 have three distinct roots, namely α, β and γ. Then, each of the three roots must satisfy the equation.

∴ aα^{2} + bα + c = 0 ……….(i)aβ ^{2} + bβ + c = 0 ……….(ii)and aγ ^{2} + bγ + c = 0 ……….(iii)Subtracting (ii) from (i) and (iii) from (ii), we get a(α ^{2} – β^{2}) + b(α – β) = 0 ……….(iv)and a(β ^{2} – γ^{2}) + b(β – γ) = 0 ……….(v)From (iv), a(α – β)(α + β) + b(α – β) = 0 ⇒ (α – β) {a(α + β) + b} = 0 α ≠ β, as the roots are distinct. ∴ a(α + β) + b = 0 ………(vi) Similarly, from (v), we get a(β + γ) + b = 0 ………(vii) Subtracting (vii) from (vi), we get a(α – γ) = 0 But a ≠ 0 ∴ α = γ, which contradicts our assumption that the roots are distinct. Thus, a quadratic equation cannot have more than two roots. |

(2)* Irrational roots of a quadratic equation with rational coefficients occur in conjugate pair.*

**Proof:** Let α + √β, where α and β are rational, be an irrational root of the quadratic equation ax^{2} + bx + c = 0, where a, b, c are rational and a ≠ 0. Then, α + √β must satisfy the equation.

∴ a(α + √β)^{2} + b(α + √β) + c = 0⇒ a(α ^{2} + 2α√β + β) + bα + b√β + c= 0⇒ (aα ^{2} + aβ + bα + c) + (2aα + b)√β = 0Now, equating the rational and irrational parts separately, we get aα ^{2} + aβ + bα + c = 0and (2aα + b) = 0 ∴ aα ^{2} + aβ + bα + c – (2aα + b)√β = 0⇒ a(α ^{2} – 2α√β + β) + b(α – √β) + c = 0⇒ a(α – √β) ^{2} + b(α – √β) + c = 0, which implies that α – √β also satisfies the quadratic equation.Thus, if α + √β is a root of the quadratic equation ax ^{2} + bx + c = 0, where a, b, c, α and β are all rational, then α – √β must also be another root. As a quadratic equation has only two roots, α + √β and α – √β are the two roots, which occur in conjugate pair in the above case. |

(3) *Imaginary roots of a quadratic equation with real coefficients must occur in conjugate pair.*

**Proof:** Let α + *i*β, with α, β as real, be an imaginary root of the quadratic equation ax^{2} + bx + c = 0 with a, b, c real and α ≠ 0. Then α + *i*β must satisfy the quadratic equation.

∴ a(α + iβ)^{2} + b(α + iβ) + c = 0⇒ aα ^{2} + 2aαβi – aβ^{2} + bα + bβi + c = 0⇒ (aα ^{2} – aβ^{2} + bα + c) + (2aαβ + bβ)i = 0Now, equating the real and imaginary parts separately, we get aα ^{2} – aβ^{2} + bα + c = 0and 2aαβ + bβ = 0 ∴ (aα ^{2} – aβ^{2} + bα + c) – (2aαβ + bβ)i = 0⇒ a(α ^{2} – 2αβi – β^{2}) + b(α – iβ) + c = 0⇒ a(α – iβ)^{2} + b(α – iβ) + c = 0, which implies that α – iβ also satisfies the given equation.Thus, if α + iβ is a root of the quadratic equation with real coefficients, then α – iβ must be the other root.Hence, imaginary roots of a quadratic equation with real coefficients occur in conjugate pair. |