84 comparison test
Completing the square
comparison test See
CONVERGENT SERIES
.
completing the square A
QUADRATIC
quantity of the
form x2+ 2bx can be regarded, geometrically, as the
formula for the area of an incomplete square.
Adding the term b2completes the picture of an
(x+b) ×(x+ b) square. We have:
x2+ 2bx + b2= (x+ b)2
This process of completing the square provides a
useful technique for solving quadratic equations. For
example, consider the equation x2+ 6x+ 5 = 21. Com-
pleting the square of the portion x2+ 6xrequires the
addition of the constant term 9. We can achieve this by
adding 4 to both sides of the equation. We obtain:
x2+ 6x+ 5 + 4 = 21 + 4
x2+ 6x+ 9 = 25
(x+ 3)2= 25
from which it follows that x+ 3 equals either 5 or –5,
that is, that xequals 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 ax2+ 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
aand add a term to complete the square of resultant
portion . We have:
For example, to solve x2+ 6x+ 5 = 21, subtract 21 from
both sides of the equation to obtain x2+ 6x– 16 = 0. By
the quadratic formula:
The quadratic formula shows that the two roots r1
and r2of a quadratic equation ax2+ bx + c= 0 (or
the single double root if the
DISCRIMINANT
b2–4ac
equals zero) satisfy and . It also
shows that every quadratic equation can be solved if
one is willing to permit
COMPLEX NUMBERS
as solu-
tions. (One may be required to take the square root of
a negative quantity.)
There do exist analogous formulae for solving
CUBIC EQUATION
s ax3+ bx2+ cx + d= 0 and
QUARTIC
EQUATION
s ax4+ bx3+ cx2+ dx + e= 0 in terms of the
coefficients that appear in the equations. Algebraist
N
IELS
H
ENRIK
A
BEL
(1802–29) showed that there can
be no analogous formulae for solving fifth- and higher-
degree equations.
See also
FACTORIZATION
;
FUNDAMENTAL THEOREM
OF ARITHMETIC
;
HISTORY OF EQUATIONS AND ALGEBRA
(essay);
SOLUTION BY RADICALS
.
rr c
a
12=
rr b
a
12
+=−
x=−± − ⋅− =−± =−± =−
636416
2
6 100
2
610
228
() or
xb
axc
a
xb
axb
a
c
a
b
a
xb
a
c
a
b
a
xb
a
b
a
c
a
bac
a
xb
a
bac
a
xbb ac
2
2
22
22
2
22
2
2
2
2
2
0
22
24
24
4
4
2
4
2
4
++=
++
+=
+
+=
+
=−=
−
+=± −
=−± −
22a
xb
ax
2+
xbb ac
a
=−± −
24
2