1.Algebra Booster

2.40 Algebra Booster
117. The given equation is
2x 4 – x 3 – 11x 2 – x + 2 = 0
Ê 2 1 ˆ Ê
2 x x
1 ˆ
Á + - + - 11=
0
Ë 2 ˜ Á ˜
x ¯ Ë x¯
1
Put x+ = t
x
Then the given equation reduces to
2(t 2 – 2) – t – 11 = 0
2t 2 – t – 15 = 0
(t – 3)(2t + 5) = 0
5
t = 3, -
2
1 5
x + = 3, -
x 2
x 2 – 3x + 1 = 0, 2x 2 + 5x + 2 = 0
3±
5 1
x= , x=-2,
-
2 2
Hence, the solution set is
Ï 1 3±
5¸
Ì-2, - , ˝
Ó 2 2 ˛
118. The given equation is
(x – 1)(x – 1)(x – 1)(x – 1) = 8
{(x – 1)(x – 4){((x – 2)(x – 3)} = 8
{(x 2 – 5x + 4)}{x 2 – 5x + 6)} = 8
(a + 4)(a + 6) = 8, x 2 – 5x = a
(a + 4)(a + 6) = 8
a 2 + 10a + 16 = 0
(a + 2)(a + 8) = 0
a = –2, –8
x 2 – 5x = –2, –8
x 2 – 5x + 2 = 0, x 4 – 5x + 8 = 0
5± 17 5±
i 7
x = ,
2 2
119. The given equation is
2
2 x
x + =
( x + 1)
2
3
x 4 + 2x 3 – x 2 – 6x – 3 = 0
Let f(x) = x 4 + 2x 3 – x 2 – 6x – 3
f(x): + + – – –
f(–x): + – – + –
In f(x), there are one change, so f(x) has one +ve root
and in f(–x), there are 3 changes, so f(–x) has 3 negative
roots
Thus, the number of real roots = 1 + 3 = 4
120. Replace x by 1 we get
x
32x 3 – 48x 2 + 22x – 3 = 0
Let its roots are a – b, a, a + b
Sum of the roots = 48 =
3
32 2
3
3a
=
2
1
a =
2
3
Also, aa ( + b)( a– b)
=
32
2 2 3
aa ( - b ) =
32
1Ê1 2ˆ
3
Á - b =
2Ë
˜
4 ¯ 32
1
b =±
4
Hence, the roots are
1 1 1 1 1 1 1 1 1 1
- , , + or + , , -
2 4 2 2 4 2 4 2 2 4
1 1 3 3 1 1
, , or , ,
4 2 4 4 2 4
Therefore, the roots of the main equation are
4 4
4, 2, or , 2, 4
3 3
121. The given equation is
6 – 11x + 6x 2 – x 3 = 0
Replace x by 1/x, we get
x 3 – 6x 2 + 11x – 6 = 0
(x – 1)(x – 2)(x – 3) = 0
x = 1, 2, 3
1 1
Hence, the solutions are 1, ,
2 3
122. Let y = ab + ag + bg – bg
y = q – bg
abg r
y = q - = q +
a a
r
a =
y - q
Since a is a root of the given equation, so
a 3 + pa 2 + qa + r = 0
3 2
Ê r ˆ Ê r ˆ Ê r ˆ
Á + p + q + r = 0
Ë y – q˜ ¯
Á
Ë y – q˜ ¯
Á
Ë y – q˜
¯
Hence, the required equation is
r 3 + pr(x – q) + q(x – q) 2 + r(x – q) 3 = 0
b + g a + b + g -a
p
123. Let y = = = -1
a a a
p
a =
y + 1