Quadratic Equation cannot have more than Two Roots

We will discuss here that a quadratic equation cannot have more than two roots.

Proof:

Let us assumed that α, β and γ be three distinct roots of the quadratic equation of the general form ax$^{2}$ + bx + c = 0, where a, b, c are three real numbers and a ≠ 0. Then, each one of α, β and γ will satisfy the given equation ax$^{2}$ + bx + c = 0.

Therefore,

aα$^{2}$ + bα + c = 0 ............... (i)

aβ$^{2}$ + bβ + c = 0 ............... (ii)

aγ$^{2}$ + bγ + c = 0 ............... (iii)

Subtracting (ii) from (i), we get

a(α$^{2}$ - β$^{2}$) + b(α - β) = 0

⇒ (α - β)[a(α + β) + b] = 0

⇒ a(α + β) + b = 0, ............... (iv) [Since, α and β are distinct, Therefore, (α - β) ≠ 0]

Similarly, Subtracting (iii) from (ii), we get

a(β$^{2}$ - γ$^{2}$) + b(β - γ) = 0

⇒ (β - γ)[a(β + γ) + b] = 0

⇒ a(β + γ) + b = 0, ............... (v) [Since, β and γ are distinct, Therefore, (β - γ) ≠ 0]

Again subtracting (v) from (iv), we get

a(α - γ) = 0

⇒ either a = 0 or, (α - γ) = 0

But this is not possible, because by the hypothesis a ≠ 0 and α - γ ≠ 0 since α ≠ γ

α and γ are distinct.

Thus, a(α - γ) = 0 cannot be true.

Therefore, our assumption that a quadratic equation has three distinct real roots is wrong.

Hence, every quadratic equation cannot have more than 2 roots.

Note: If a condition in the form of a quadratic equation is satisfied by more than two values of the unknown then the condition represents an identity.

Consider the quadratic equation of the general from ax$^{2}$ + bx + c = 0 (a ≠ 0) ............... (i)

Solved examples to find that a quadratic equation cannot have more than two distinct roots

Solve the quadratic equation 3x$^{2}$ - 4x - 4 = 0 by using the general expressions for the roots of a quadratic equation.

Solution:

The given equation is 3x$^{2}$ - 4x - 4 = 0

Comparing the given equation with the general form of the quadratic equation ax^2 + bx + c = 0, we get

a = 3; b = -4 and c = -4

Substituting the values of a, b and c in α = $\frac{- b - \sqrt{b^{2} - 4ac}}{2a}$ and β = $\frac{- b + \sqrt{b^{2} - 4ac}}{2a}$ we get

α = $\frac{- (-4) - \sqrt{(-4)^{2} - 4(3)(-4)}}{2(3)}$ and β = $\frac{- (-4) + \sqrt{(-4)^{2} - 4(3)(-4)}}{2(3)}$

⇒ α = $\frac{4 - \sqrt{16 + 48}}{6}$ and β =$\frac{4 + \sqrt{16 + 48}}{6}$

⇒ α = $\frac{4 - \sqrt{64}}{6}$ and β =$\frac{4 + \sqrt{64}}{6}$

⇒ α = $\frac{4 - 8}{6}$ and β =$\frac{4 + 8}{6}$

⇒ α = $\frac{-4}{6}$ and β =$\frac{12}{6}$

⇒ α = -$\frac{2}{3}$ and β = 2

Therefore, the roots of the given quadratic equation are 2 and -$\frac{2}{3}$.

Hence, a quadratic equation cannot have more than two distinct roots.

11 and 12 Grade Math 

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