Example
1
Consider the
inequation x2 + |x + a| –9 < 0, find the values of the real
parameter 'a' so that the given inequation has atleast one negative solution.
Solution:
|
9 – x2 > |x + a|
As
per the question, we have to make sure that for atleast one negative x, graph
of y = 9 – x2 must lie above the graph of y = |x + a|
Case-I:
a ≥ 0, Let us take up the limiting case when y= |x + a|
touches the parabola
y
= 9 – x2
&⇒ 9 – x2 = x + a should
have equal roots
&⇒
x2 + x + (a – 9) = 0 has equal roots
&⇒
1 – 4(a – 9) = 0
&⇒ 
&⇒
0 £ a < 
|
|
|
Case-II:
a < 0
In the limiting case the left
branch of y = | x + a |
i.e. y = – x – a will pass
through the vertex of the parabola.
&⇒
9 = 0 – a
&⇒ a = –9
&⇒
a > – 9
That
means required set of values of a is 
|
|
Alternative
Solution:
x2 + |x +a| - 9 < 0
Case –I Let x +a ≥
0
&⇒ For all x < 0, a > 0
Also x2 +x +a – 9 < 0
&⇒
a < 9 – x2 –x
&⇒
a < 
&⇒
a <
i.e. 0 < a < . . .
(1)
Case –II Let x +a < 0
&⇒
a£
0
Also x2 –x –a –9 < 0
&⇒
a > x2 –x – 9
&⇒
a > - 9 ,
so, -9 < a £
0 . . . . (2).
From (1) and (2), -9 < a < .
Example
2
Show
that if x is real, the expression 
has no value lying
between b and c.
Solution:
Let 
&⇒ x2
– 2xy + (b + c)y – bc = 0
Now x is real
&⇒
Discriminant ≥ 0
&⇒
4y2 – 4[(b + c)y – bc] ≥ 0
&⇒ (y – b) (y –
c) ≥ 0
&⇒
y has no value between b and c
Example
3
Solve the equation log4 (2x2
+ x + 1) – log2 (2x –1) = 1.
Solution:
log4(2x2 + x +
1) – log2(2x –1) = 1
&⇒
&⇒
loge
&⇒
(2x2 + x + 1) = 4(4x2 – 4x + 1)
&⇒
14 x2 - 17x + 3 = 0
&⇒ (14x - 3) (x - 1) = 0
&⇒
x = 
, x = 1
But x = does not lies in the domain
of function.
Hence x = 1 is only solution. .
Example
4
Let a,b,c be real.
If ax2+bx+c=0 has two real roots a
and b where a < - 1 and b > 1, then show that 
Solution:
a
< -1
&⇒
a
+ E1 = -1, where E1 > 0
Also b
> 1
&⇒
b
- E2 = 1, where E2 > 0.
= 1-1-E1-E2-E1E2+½E2-E1½ =
-E1-E2-E1E2+E2-E1
, if E2 > E1
= -E1-E2-E1E2+E1-E2 if
E1 > E2. .
Hence L.H.S. = -2E1-E1E2
or -2E2-E1E2.
In both the cases 
. (E1,
E2 > 0)
|
Alternative Method
Let f(x) = x2 +  x + 
|
|
from graph f(-1) < 0 and f(1)
< 0
&⇒
1+
< 0 and 1 + + < 0
&⇒
1 + + 
< 0.
Example
5
Find all the real
values of the parameter a for which the equation x(x+1)(x+a)(x+a+1)=a2
has four real roots.
Solution:
Rewritting the given equation as (x2
+ax + x )(x2+ax + x+a) = a2 and put x2 + ax +
x= y, so that y(y +a) = a2
or y2 + ay - a2
= 0 whose solution is
y = 
There will then be four real roots
if each of the following two equations x2 + x(a + 1) = 
a has two real roots which implies
(a +1)2 + 4
&⇒ a2
+2a +1 –2a ± 2
a ≥ 0
&⇒
a2 ± 2a +1 ≥ 0
&⇒
(a + )2 ≥
4 and (a -)2 ≥
4
This leads to the four
possibilities
a +£ -2, a +≥ 2, a - £
-2 or a - ≥ 2
In other words, a £
-2 - , a ≥ 2 - , a £ -2 or a ≥
+2 .
It is clear that 2 - £ a £
-2 , a £
-2 -or a ≥ 2 +
Whence |a| £
-2 or |a| ≥
2 + .
