Question

prove that any quadratic equation has maximum two root

Answer 1

Proof:

Let us assumed that α, β and γ be three distinct roots of the quadratic equation of the general form ax

2

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

2

+ bx + c = 0.

Therefore,

aα

2

2

+ bα + c = 0 ............... (i)

aβ

2

2

+ bβ + c = 0 ............... (ii)

aγ

2

2

+ bγ + c = 0 ............... (iii)

Subtracting (ii) from (i), we get

a(α

2

2

- β

2

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

2

- γ

2

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.