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
From Quadratic Equation cannot have more than Two Roots to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Sep 15, 24 04:57 PM
Sep 15, 24 04:08 PM
Sep 15, 24 03:16 PM
Sep 14, 24 04:31 PM
Sep 14, 24 03:39 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.