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Fundamental Theorem Of Algebra For Quadratic Equations

Fundamental Theorem Quadratic Equation Online Tutoring / Homework Help
STATEMENT:

Every algebraic equation with real or imaginary coefficients has at least one root, real or imaginary.

THEOREM:

A quadratic equation cannot have more than two distinct roots.

Proof:

Let a quadratic equation be ax2 + bx + c = 0 [a ≠ 0]             (1)

Then by the fundamental theorem of algebra the equation (1) has at least one root. Let this root be Alpha. Then, by the remainder theorem (x - Alpha) is a factor of the expression ax2 + bx + c and aAlpha 2 + bAlpha + c = 0. Hence,

ax2 + bx + c = (x - Alpha)(px + q)             (2)

where p ( ≠ 0) and q are constants [Since quotient is a linear expression when the quadratic expression (ax2 + bx + c) is divided by (x - Alpha)].

Again, by fundamental theorem the equation px + q = 0 has at least one root. Let this root be β. Then, (x - β) is a factor of px + q = 0 and pβ + q = 0. Hence, we get,

px + q = (x - β).r             (3)

where r is a constant[Since quotient is a constant when (px + q) is divided by (x - β)]

Thus, from (2) and (3) we get,

ax2 + bx + c = r(x - Alpha)(x - β)             (4)

Equating the coefficients of x2 on both sides of (4) we get, r = a.

Hence, ax2 + bx + c = a(x - Alpha)(x - β)             (5)

From (5) it is clear that ax2 + bx + c = 0 when x = Alpha or x = β. For this reason, x = Alpha and x = β, both are roots of equation (1) that is, the quadratic equation (1) has two roots Alpha and β.

Now, if we take a quantity γ such that γ ≠ Alpha and γ ≠ β. Hence, γ - Alpha ≠ 0 and γ - β ≠ 0. Thus, by putting x = γ on both sides of (5) we get,aγ 2 + bγ + c = a(γ - Alpha) ( γ - β) ≠ 0 [Since a ≠ 0]

Hence, x = γ cannot be a root of equation (1).

Thus, a quadratic equation cannot have more than two distinct roots. (Proved)


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