Quadratic Equation
Basic Concepts:
An equation of the form ax2
+ bx + c = 0, where a > 0 and a, b, c are real numbers, is
called a quadratic equation.
The numbers a, b, c are called the
coefficients of the quadratic equation. A root of the quadratic equation is a
number a (real or complex) such that aa2 + ba + c = 0.
The roots of the quadratic equation
are given by
x = 
The quantity D (D= b2 -
4ac) is known as the discriminant of the equation.
Basic Results:
- The quadratic equation has real and
equal roots if and only if D = 0 i.e. b2- 4ac=0.
- The quadratic equation has real and
distinct roots if and only if D > 0 i.e. b2-4ac>0.
- The quadratic equation has complex
roots with non-zero imaginary parts if and only if D < 0 i.e. b2
- 4ac < 0.
- If p + iq (p and q being real) is a
root of the quadratic equation where i =
, then p - iq
is also a root of the quadratic equation. - If p +
is an
irrational root of the quadratic equation, then p - is also a root
of the quadratic equation provided that all the coefficients are rational, q
not being a perfect square or zero. - The quadratic equation has rational
roots if D is a perfect square and a, b, c are rational.
- If a = 1 and b, c are integers and
the roots of the quadratic equation are rational, then the roots must be
integers.
- If the quadratic equation is
satisfied by more than two distinct numbers (real or complex), then it becomes
an identity i.e. a = b = c = 0.
- Let a and b
be two roots of the given quadratic equation. Then a
+ b = -
and ab = 
. - A quadratic equation, whose roots are
a
and b can be written as (x - a) (x - b)
= 0 i.e., ax2 + bx + c º a(x - a) (x - b).
Example 1
Prove that
the roots of the quadratic equation ax2 - 3bx - 4a = 0 are real and
distinct for all real a and b. .
Solution:
D = (-3b)2
– 4(-4a) (a)
= 9b2 + 16a2
which is always positive (as a > 0). .
Hence the roots will be
real and distinct.
Example 2
Prove that the roots
of ax2 + 2bx + c = 0 will be real and distinct if and only if the
roots of (a + c) (ax2 + 2bx + c) = 2(ac-b2) (x2
+1) are imaginary. .
Solution:
Let D1 be the discriminant
of ax2 +2bx +c = 0 … (1) The other equation
(a + c) (ax2 +2bx +c) = 2(ac -b2) (x2 +1) can
be
written as (a2 +2b2
–ac ) x2+2(ab+bc) x + (c2 + 2b2 –ac) = 0 …(2)
Hence D1 = 4b2
- 4ac = 4 (b2 -ac) and
D2 = 4(ab +bc)2
– 4( c2 +2b2 –ac) (a2 +2b2 –ac)
= 4 (-a2b2 – b2
c2 – 2a2c2 + 6ab2c-4b4
+ac3 +a3c)
= - 4[ 4b2(b2 –
ac) +a2(b2 – ac)+ c2(b2 – ac) –
2ac(b2 – ac) ]
= - 4(b2-ac)(4b2+(a-c)2)
Since D2 < 0 Û
b2 - ac > 0
or, D2 < 0 Û
D1 > 0,
the roots of (1) are real and
distinct if and only if the roots of (2) are imaginary.
Example 3
Let a and b be the roots of the equation ax2 + 2bx +
c = 0 and a + g and b + g
be the roots of Ax2 +2Bx +C = 0. Then prove that
A2(b2 - ac) = a2(B2 - AC).
Solution:
For the given equation, a
+b
= -
, ab =
and a
+b
+2g = 
,
(a+g)
(b+g)
= 
&⇒
2g
= 
, ab +g(a+b)
+g2
=
&⇒
g
= 
, ab + g(a
+ b + g) =
Eliminating a,
b
and g, we get
&⇒ 
&⇒
a2 (B2-AC) = A2(b2-ac).
Alternative:
Let f(x) = x2
+ 
, g(x) = x2
+
Since roots of f(x) = 0
is a and b and that of g(x) = 0 are a
+ g
and b + g
&⇒ Graph of g(x)
will be obtained from the graph of f(x) by translating it g
unit on the x-axis.
Therefore minimum of
f(x) = minimum of g(x)
&⇒
&⇒
&⇒
A2 (b2 –ac)= a2 (B2 –AC).
Example 4
If a,
b and c are odd integers, then prove that roots of ax2 + bx + c = 0
can not be rational.
Solution:
Discriminant D = b2 –4ac. .
Suppose the roots are rational. Then
D will be a perfect square. .
Let b2 - 4ac = d2.
Since a, b and c are odd integers, d will be odd. .
Now b2 - d2 =
4ac.
Since b and d are odd integers, b2
– d2 will be a multiple of 8
b2 – d2 = (b
–d) (b + d), say b = 2k + 1 and d = 2m + 1
&⇒
b2 –d2 = 2(k –m)2(k + m + 1)
now either (k –m) or (k + m + 1) is
always even hence b2 – d2 is always a multiple of 8. .
But 4ac is
only a multiple of 4 (not of 8), which is a contradiction. Hence the roots of
ax2 + bx + c = 0 can not be rational.
