Example
1
Consider the
inequation x^{2} + x + a –9 < 0, find the values of the real
parameter 'a' so that the given inequation has atleast one negative solution.
Solution:
9 – x^{2} > x + a
As
per the question, we have to make sure that for atleast one negative x, graph
of y = 9 – x^{2} must lie above the graph of y = x + a
CaseI:
a ≥ 0, Let us take up the limiting case when y= x + a
touches the parabola
y
= 9 – x^{2 } &⇒ 9 – x^{2} = x + a should
have equal roots
&⇒
x^{2} + x + (a – 9) = 0 has equal roots
&⇒
1 – 4(a – 9) = 0 &⇒
&⇒
0 £ a <


CaseII:
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:
x^{2} + x +a  9 < 0
Case –I Let x +a ≥
0
&⇒ For all x < 0, a > 0
Also x^{2} +x +a – 9 < 0
&⇒
a < 9 – x^{2} –x
&⇒
a <
&⇒
a <
i.e. 0 < a < . . .
(1)
Case –II Let x +a < 0
&⇒
a£
0
Also x^{2} –x –a –9 < 0
&⇒
a > x^{2} –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
&⇒ x^{2}
– 2xy + (b + c)y – bc = 0
Now x is real
&⇒
Discriminant ≥ 0
&⇒
4y^{2} – 4[(b + c)y – bc] ≥ 0
&⇒ (y – b) (y –
c) ≥ 0
&⇒
y has no value between b and c
Example
3
Solve the equation log_{4 }(2x^{2
}+ x + 1) – log_{2 }(2x –1) = 1.
Solution:
log_{4}(2x^{2} + x +
1) – log_{2}(2x –1) = 1
&⇒
&⇒
log_{e}_{}
&⇒
(2x^{2} + x + 1) = 4(4x^{2} – 4x + 1)
&⇒
14 x^{2}  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 ax^{2}+bx+c=0 has two real roots a
and b where a <  1 and b > 1, then show that
Solution:
a
< 1
&⇒
a
+ E_{1} = 1, where E_{1} > 0
Also b
> 1
&⇒
b
 E_{2 }= 1, where E_{2} > 0.
= 11E_{1}E_{2}E_{1}E_{2}+½E_{2}E_{1}½ =
E_{1}E_{2}E_{1}E_{2}+E_{2}E_{1}
, if E_{2} > E_{1}
= E_{1}E_{2}E_{1}E_{2}+E_{1}E_{2 }if
E_{1 }> E_{2}. .
Hence L.H.S. = 2E_{1}E_{1}E_{2}
or 2E_{2}E_{1}E_{2}.
In both the cases . (E_{1},
E_{2} > 0)
Alternative Method
Let f(x) = x^{2} + 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)=a^{2}
has four real roots.
Solution:
Rewritting the given equation as (x^{2}
+ax + x )(x^{2}+ax + x+a) = a^{2} and put x^{2} + ax +
x= y, so that y(y +a) = a^{2}
or y^{2} + ay  a^{2}
= 0 whose solution is
y =
There will then be four real roots
if each of the following two equations x^{2} + x(a + 1) = a has two real roots which implies
(a +1)^{2} + 4
&⇒ a^{2}
+2a +1 –2a ± 2a ≥ 0
&⇒
a^{2} ± 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 + .