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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