Example
1
If a, b
are the roots of x2+px+q=0, and g,d are the roots of x2 + rx
+ s = 0, evaluate (a – g) (a – d)
(b – g) (b – d)
in terms of p, q, r and s.
Deduce
the condition that the equations have a common root.
Solution:
We have a+b=-p,
ab=q, g+d=
- r, gd=s
Hence (a-g)
(a-d)
(b-g)
(b-d)
= [a2-a(g+d)+
gd][b2- b(g+d)+
gd]
= [a2+ra+s][b2+rb+s] = a2b2+rab(a+b)+r2ab+s[a2+b2]+s2+sr(a+b)
= q2-rpq + r2q
+ s(p2-2q)+s2-srp = (q-s)2 + (p-r) (sp-rq)
For a common root
R.H.S. = 0
&⇒ (q-s)2 = (p-r) (rq-sp),
which is the required condition.
Example
2
Find all the real
numbers x satisfying
Solution:
Let
= A
and 
= B
We have A + B = 
and AB = 
Therefore (A - B)2 = (A +
B)2 - 4AB
= 4. - 4. = 0
&⇒ A = B
Since powers in A and B are same,
either the power is equal to zero or the bases are the same.
&⇒
x = 0 or 
&⇒
x = 0 or 
=0
&⇒ x = 0 or x
= 4 or 1.
Example
3
Let
x, y, z be real variables satisfying the equations x+y +z = 6 and
xy + yz +zx = 7. Then find the range in which the variables
can lie.
Solution:
We are given that
x+ y+ z = 6 . .
. . (1).
and xy +yz +zx =7 .
. . . . (2) .
From (1), z = 6 - x -y
Putting the value of z in (2),
we get
xy+ y(6-x-y) + x( 6-x-y) =7
or, y2 +y(x-6) + x2
–6x +7 =0
Since y is real , (x-6)2
- 4(x2 –6x + 7 ) ≥ 0
or, 3x2 - 12x –8 £
0
&⇒ 
Since (1) and (2) are
symmetrical in x, y and z, all the variable lie in the interval 
.
Example
4
If one root of the
quadratic equation ax2 + bx + c = 0 is equal to nth ( n is an even
natural number) power of the other root, then show that 
.
Solution:
Let a be one of the
roots. Then the other root is an.
Hence a+an = -b/a and an+1
= c/a
.
Example
5
Find all the
values of the parameter c for which the inequality
1 + log2(2x2 +2x +
) ≥ log2(cx2
+c) possesses at least one solution.
Solution:
Given inequality is 1 + log2(2x2
+2x +
) ≥ log2(cx2
+c) ….. (1)
In order that (1) make sense we must
have c> 0. Also .
2x2 + 2x += 2(x2 + x + 
) = 2
" x ∈
R
We can write (1) as log2
&⇒
cx2 + c £ 4x2+ 4x + 7
&⇒
(c - 4)x2 - 4x +c - 7£ 0 …..(2).
We have D = b2 - 4ac = 16
- 4(c - 4)(c - 7)
= -4(c2 - 11c + 24) = -4(c
- 3)(c - 8)
Case-I : If D < 0, that is, if
c < 3 or c > 8 and c - 4< 0 then(2) holds for all x∈
R. Thus if 0 < c < 3. Then (1) holds for
all x ∈ R. .
Case-II: If D ≥
0, that is, if 3 £ c £ 8 and c –
4< 0
then (2) holds for all x £
-(4 + 
)/2(4 - c) and for all
x ≥ -(4 - )/2(4 - c). Thus,
if 3 £ c < 4 then
there is at least one x satisfying (1). .
Case-III: If D ≥
0, that is, if 3 £ c £ 8 and c - 4
> 0 then (2) holds for all x lying between -(4+)/2(4 - c) and -(4 +)/2(4 -c). Thus for 4 < c £
8 there is at least one x satisfying (1). Also for c = 4, (2) is satisfied
for all x ≥ -3/4
Hence if 0 < c £
8, there is at least one x satisfying the inequality (1)
