Question

(iv) 3x - 24-X
х
3 x=0
I​

Answer 1

120=(+1)(+2)(+3)(+4)=((+1)(+4))((+2)(+3))
120
=
(
x
+
1
)
(
x
+
2
)
(
x
+
3
)
(
x
+
4
)
=
(
(
x
+
1
)
(
x
+
4
)
)
(
(
x
+
2
)
(
x
+
3
)
)


=((2+5+5)−1)((2+5+5)+1)…(1)
=
(
(
x
2
+
5
x
+
5
)
−
1
)
(
(
x
2
+
5
x
+
5
)
+
1
)
…
(
1
)


=(2+5+5)2−1
=
(
x
2
+
5
x
+
5
)
2
−
1
,

we get 2+5+5=±11
x
2
+
5
x
+
5
=
±
11
.

Thus 2+5−6=0
x
2
+
5
x
−
6
=
0
or 2+5+16=0
x
2
+
5
x
+
16
=
0
. The first of these give =1
x
=
1
or =−6
x
=
−
6
. The second of these give =12(−5±−39‾‾‾‾√)
x
=
1
2
(
−
5
±
−
39
)
.

It is easy to verify that =1
x
=
1
and =−6
x
=
−
6
are both solutions; in fact, 120=2⋅3⋅4⋅5=(−5)⋅(−4)⋅(−3)⋅(−2)
120
=
2
⋅
3
⋅
4
⋅
5
=
(
−
5
)
⋅
(
−
4
)
⋅
(
−
3
)
⋅
(
−
2
)
.

That the second pair of complex conjugates is also a solution can be best verified by replacing 2+5
x
2
+
5
x
by −16
−
16
in eqn. (1)
(
1
)
, say.

There are two real solutions, =1
x
=
1
and =−6
x
=
−
6
, and two non-real solutions =12(−5±−39‾‾‾‾√)
x
=
1
2
(
−
5
±
−
39
)
.