Expressions Reducible to Quadratic Form
Involving variable as
powers :
In these questions a particular
number with some variable power is substituted as a new variable, in order to
make the equation quadratic in new variable. .
Example 1
Solve 
Solution:
= z
&⇒ 
&⇒
z – 
=
1 = 0
&⇒ z = 2 or z = –1
&⇒
( z – 2)(z + 1) = 0
&⇒ z = 2 or z = 1
&⇒
=
2 or =
–1 (not possible)
&⇒
sin2x = 1
&⇒ x = (2n + 1) 
.
Example 2
Solve 4x + 6x = 9x. .
Solution:
4x + 6x = 9x
&⇒
Put
&⇒
y2 + y – 1 = 0
&⇒
y = 
&⇒
&⇒
x = 
.
Involving logarithms:
By using basic definition of
logarithm i.e. logab = x
&⇒ ax = b "
a, b, > 0 and a > 1 .
We change logarithmic equation to
exponential and proceed as in 8.1. .
If equation or inequation compares
two different logarithmic expressions then if they have the same base, we use
following basic theorem
i.e. logax > logay
&⇒
x > y if a > 1
&⇒ x < y if a < 1 .
Example 3
Solve log 3(3x – 8) = 2 – x
Solution:
log3(3x – 8) =
log3 (32 –x)
&⇒ 3x – 8 = 32–x
= 
put 3x= z
z – 8 = 
&⇒
z2 – 8z – 9 = 0
&⇒
(z – 9 ) ( z+ 1) = 0
&⇒ z = 9 and z = –1 (not
possible)
z = 9
&⇒ x = 2.
Example 4
Solve log9(x2 – 5x +
6) > log3( x– 4)
Solution:
&⇒
x2 – 5x + 6 > x2 – 8x + 16
&⇒
3x – 10 > 0
&⇒ x > 
……(1)
also, x2 – 5x + 6 > 0
&⇒
x > 3 or x < 2 ……(2)
and x – 4 > 0
&⇒
x > 4 ……(3)
common solution from (1), (2) and (3)
x > 4.
Involving Modulus:
By using the basic definition of
modulus i.e. 
To find x such that f(x) is > or
< 0 we use wavy curve method if f(x) can be factorised, or any suitable
method
Example 5
Solve |x – 3| + |x – 7| > 8
Solution:
Case (i) x < 3
&⇒
–x + 3 – x + 7 > 8
&⇒ – 2x > –2
&⇒
x < 1
Case (ii) 3 £
x < 7
&⇒ x – 3 – x + 7 > 8
no such exists
Case(iii) x ≥
7
&⇒ x – 3 + x – 7 > 8
&⇒
2x > 18
&⇒ x > 9
Hence x ∈
( –¥, 1) È(9, ¥)
Note:
If |f(x) + g(x) + h(x)| = |f(x)| +
|g(x)| + |h(x)|
If f(x) , g(x), h(x) are all
non-positive or all non-negative
Example 6
Solve |x2 – 3| + |2 –x| = |x2 – x
– 1|
Solution:
This is true "
x which satisfy
(x2 – 3) ( 2 – x) ≥
0
&⇒
x ∈ (–¥, –Ö3] È
[Ö3,
2]
