comparison test See CONVERGENT SERIES.
completing the square A QUADRATIC quantity of the
form x
2 + 2bx can be regarded, geometrically, as the
formula for the area of an incomplete square.
Adding the term b
2 completes the picture of an
(x + b) × (x + b) square. We have:
x
2 + 2bx + b
2 = (x + b)
2
This process of completing the square provides a
useful technique for solving quadratic equations. For
example, consider the equation x
2 + 6x + 5 = 21. Completing the square of the portion x
2 + 6x requires the
addition of the constant term 9. We can achieve this by
adding 4 to both sides of the equation. We obtain:
x
2 + 6x + 5 + 4 = 21 + 4
x
2 + 6x + 9 = 25
(x + 3)
2 = 25
from which it follows that x + 3 equals either 5 or –5,
that is, that x equals 2 or –8.
The process of completing the square generates a
general formula for solving all quadratic equations.
We have:
The solutions of a quadratic equation ax
2 + bx
+ c = 0, with a ≠ 0, are given by:
This formula is known as the quadratic formula. To see
why it is correct, divide the given equation through by
a and add a term to complete the square of resultant
portion
. We have:
For example, to solve x
2 + 6x + 5 = 21, subtract 21 from
both sides of the equation to obtain x
2 + 6x – 16 = 0. By
the quadratic formula:
The quadratic formula shows that the two roots r 1
and r 2 of a quadratic equation ax
2 + bx + c = 0 (or
the single double root if the DISCRIMINANT b
2 –4ac
equals zero) satisfy
and
. It also
shows that every quadratic equation can be solved if
one is willing to permit COMPLEX NUMBERS as solutions. (One may be required to take the square root of
a negative quantity.)
There do exist analogous formulae for solving
CUBIC EQUATIONs ax
3 + bx
2 + cx + d = 0 and QUARTIC
EQUATIONs ax
4 + bx
3 + cx
2 + dx + e = 0 in terms of the
coefficients that appear in the equations. Algebraist
NIELS HENRIK ABEL (1802–29) showed that there can
be no analogous formulae for solving fifth- and higherdegree equations.
See also FACTORIZATION; FUNDAMENTAL THEOREM
OF ARITHMETIC; HISTORY OF EQUATIONS AND ALGEBRA
(essay); SOLUTION BY RADICALS.
r r
c
a
1 2 =
r r
b
a
1
2
+ = −
x =
− ±
− ⋅ −
=
− ±
=
− ± =
−
6
36 4 16
2
6
100
2
6 10
2
2
8
(
)
or
x
b
a
x
c
a
x
b
a
x
b
a
c
a
b
a
x
b
a
c
a
b
a
x
b
a
b
a
c
a
b
ac
a
x
b
a
b
ac
a
x
b
b
ac
2
2
2
2
2
2
2
2
2
2
2
2
2
2
0
2
2
2
4
2
4
4
4
2
4
2
4
+
+ =
+
+
+ =
+
+ =
+
=
− =
−
+
= ±
−
=
− ±
−
2 2a
x
b
a
x
2
+
x
b
b
ac
a
=
− ±
−
2
4
2
84 comparison test
Completing the square
completing the square A QUADRATIC quantity of the
form x
2 + 2bx can be regarded, geometrically, as the
formula for the area of an incomplete square.
Adding the term b
2 completes the picture of an
(x + b) × (x + b) square. We have:
x
2 + 2bx + b
2 = (x + b)
2
This process of completing the square provides a
useful technique for solving quadratic equations. For
example, consider the equation x
2 + 6x + 5 = 21. Completing the square of the portion x
2 + 6x requires the
addition of the constant term 9. We can achieve this by
adding 4 to both sides of the equation. We obtain:
x
2 + 6x + 5 + 4 = 21 + 4
x
2 + 6x + 9 = 25
(x + 3)
2 = 25
from which it follows that x + 3 equals either 5 or –5,
that is, that x equals 2 or –8.
The process of completing the square generates a
general formula for solving all quadratic equations.
We have:
The solutions of a quadratic equation ax
2 + bx
+ c = 0, with a ≠ 0, are given by:
This formula is known as the quadratic formula. To see
why it is correct, divide the given equation through by
a and add a term to complete the square of resultant
portion
. We have:
For example, to solve x
2 + 6x + 5 = 21, subtract 21 from
both sides of the equation to obtain x
2 + 6x – 16 = 0. By
the quadratic formula:
The quadratic formula shows that the two roots r 1
and r 2 of a quadratic equation ax
2 + bx + c = 0 (or
the single double root if the DISCRIMINANT b
2 –4ac
equals zero) satisfy
and
. It also
shows that every quadratic equation can be solved if
one is willing to permit COMPLEX NUMBERS as solutions. (One may be required to take the square root of
a negative quantity.)
There do exist analogous formulae for solving
CUBIC EQUATIONs ax
3 + bx
2 + cx + d = 0 and QUARTIC
EQUATIONs ax
4 + bx
3 + cx
2 + dx + e = 0 in terms of the
coefficients that appear in the equations. Algebraist
NIELS HENRIK ABEL (1802–29) showed that there can
be no analogous formulae for solving fifth- and higherdegree equations.
See also FACTORIZATION; FUNDAMENTAL THEOREM
OF ARITHMETIC; HISTORY OF EQUATIONS AND ALGEBRA
(essay); SOLUTION BY RADICALS.
r r
c
a
1 2 =
r r
b
a
1
2
+ = −
x =
− ±
− ⋅ −
=
− ±
=
− ± =
−
6
36 4 16
2
6
100
2
6 10
2
2
8
(
)
or
x
b
a
x
c
a
x
b
a
x
b
a
c
a
b
a
x
b
a
c
a
b
a
x
b
a
b
a
c
a
b
ac
a
x
b
a
b
ac
a
x
b
b
ac
2
2
2
2
2
2
2
2
2
2
2
2
2
2
0
2
2
2
4
2
4
4
4
2
4
2
4
+
+ =
+
+
+ =
+
+ =
+
=
− =
−
+
= ±
−
=
− ±
−
2 2a
x
b
a
x
2
+
x
b
b
ac
a
=
− ±
−
2
4
2
84 comparison test
Completing the square