Online Fundamental Theorem Quadratic Equation Tutor & Homework Help

Solution Fundamental Theorem Quadratic Equation Tutoring Online, Tutor Help
FUNDAMENTAL THEOREM OF ALGEBRA
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
. Then, by the remainder theorem (x -
) is a factor of
the expression ax2 + bx + c and a
2 + b
+ c = 0. Hence,
ax2 + bx + c = (x -
)(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 -
)].
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 -
)(x - β) (4)
Equating the coefficients of x2 on both sides of (4) we get, r = a.
Hence, ax2 + bx + c = a(x -
)(x - β) (5)
From (5) it is clear that ax2 + bx + c = 0 when x =
or x = β.
For this reason, x =
and x = β,
both are roots of equation (1) that is, the quadratic equation
(1) has two roots
and β.
Now, if we take a quantity γ such that γ ≠
and
γ ≠ β. Hence, γ -
≠ 0 and γ - β ≠ 0. Thus,
by putting x = γ on both sides of (5) we get,aγ 2 + bγ + c = a(γ -
)
( 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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