IIT JEE

An equation of the form ax^{2}
+ 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 aa^{2} + ba + c = 0.

The roots of the quadratic equation are given by

x =

The quantity D (D= b^{2} -
4ac) is known as the discriminant of the equation.

- The quadratic equation has real and
equal roots if and only if D = 0 i.e. b
^{2}- 4ac=0. - The quadratic equation has real and
distinct roots if and only if D > 0 i.e. b
^{2}-4ac>0. - The quadratic equation has complex
roots with non-zero imaginary parts if and only if D < 0 i.e. b
^{2}- 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., ax
^{2}+ bx + c º a(x - a) (x - b).

Prove that
the roots of the quadratic equation ax^{2} - 3bx - 4a = 0 are real and
distinct for all real a and b. .

D = (-3b)^{2}
– 4(-4a) (a)

= 9b^{2} + 16a^{2}
which is always positive (as a > 0). .

Hence the roots will be real and distinct.

Prove that the roots
of ax^{2} + 2bx + c = 0 will be real and distinct if and only if the
roots of (a + c) (ax^{2} + 2bx + c) = 2(ac-b^{2}) (x^{2}
+1) are imaginary. .

Solution:

Let D_{1} be the discriminant
of ax^{2} +2bx +c = 0 … (1) The other equation
(a + c) (ax^{2} +2bx +c) = 2(ac -b^{2}) (x^{2} +1) can
be

written as (a^{2} +2b^{2}
–ac ) x^{2}+2(ab+bc) x + (c^{2} + 2b^{2} –ac) = 0 …(2)

Hence D_{1} = 4b^{2}
- 4ac = 4 (b^{2} -ac) and

D_{2} = 4(ab +bc)^{2}
– 4( c^{2} +2b^{2} –ac) (a^{2} +2b^{2} –ac)

= 4 (-a^{2}b^{2} – b^{2}
c^{2} – 2a^{2}c^{2} + 6ab^{2}c-4b^{4}
+ac^{3} +a^{3}c)

= - 4[ 4b^{2}(b^{2 }–
ac) +a^{2}(b^{2 }– ac)+ c^{2}(b^{2 }– ac) –
2ac(b^{2 }– ac) ]

= - 4(b^{2}-ac)(4b^{2}+(a-c)^{2})

Since D_{2} < 0 Û
b^{2} - ac > 0

or, D_{2} < 0 Û
D_{1} > 0,

the roots of (1) are real and distinct if and only if the roots of (2) are imaginary.

Let a and b be the roots of the equation ax^{2} + 2bx +
c = 0 and a + g and b + g
be the roots of Ax^{2} +2Bx +C = 0. Then prove that

A^{2}(b^{2} - ac) = a^{2}(B^{2} - AC).

For the given equation, a +b = -, ab =

and a +b +2g = , (a+g) (b+g) =

&⇒
2g
= , ab +g(a+b)
+g^{2}
=

&⇒
g
= , ab + g(a
+ b + g) =

Eliminating a, b and g, we get

&⇒

&⇒
a^{2} (B^{2}-AC) = A^{2}(b^{2}-ac).

Let f(x) = x^{2}
+ , g(x) = x^{2}
+

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)

&⇒

&⇒

&⇒
A^{2} (b^{2} –ac)= a^{2} (B^{2} –AC).

If a,
b and c are odd integers, then prove that roots of ax^{2 }+ bx + c = 0
can not be rational.

Discriminant D = b^{2} –4ac. .

Suppose the roots are rational. Then D will be a perfect square. .

Let b^{2} - 4ac = d^{2}.
Since a, b and c are odd integers, d will be odd. .

Now b^{2 }- d^{2} =
4ac.

Since b and d are odd integers, b^{2}
– d^{2} will be a multiple of 8

b^{2} – d^{2} = (b
–d) (b + d), say b = 2k + 1 and d = 2m + 1

&⇒
b^{2} –d^{2} = 2(k –m)2(k + m + 1)

now either (k –m) or (k + m + 1) is
always even hence b^{2} – d^{2} 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
ax^{2 }+ bx + c = 0 can not be rational.