1.Algebra Booster

2.72 Algebra Booster
= a 2 (r 3 – 1) 2
= (ar 3 – a) 2
= (d – a) 2 = (a – d) 2
Thus, the given quadratic equation reduces to
- 9 2 2
( ) 2 0
10 x + a - d x + =
9x 2 – 10(a – d) 2 x – 20 = 0
Replace x by 1/x, we get,
2
Ê1ˆ 2 Ê1ˆ
9Á -10( a - d) - 20 = 0
Ë
˜ Á ˜
x¯ Ë x¯
9 – 10(a – d) 2 x – 20x 2 = 0
20x 2 + 10(a – d) 2 x – 9 = 0.
Hence, the result
58. Let f(x) = x 2 – 2x + a 2 + a – 3
(i) D ≥ 0
fi 4a 2 – 4(a 2 + a – 3) ≥ 0
fi a 2 – (a 2 + a – 3) ≥ 0
fi –(a – 3) ≥ 0
fi (a – 3) £ 0
fi a £ 3
(ii) af(3) > 0
fi 1.f(3) > 0
fi f(3) > 0
fi 9 – 6a + a 2 + a – 3 > 0
fi a 2 – 5a + 6 > 0
fi (a – 2)(a – 3) > 0
fi a < 2, a > 3
(iii) a + b < 2.3 = 6
fi 2a < 6
fi a < 3.
Hence, the solution set is a < 2
59. Given a + b = –b, ab = c
Since b > 0 and c < 0 , so a < 0 < b
Also, b = –b – a < –a = |a|
Therefore, a < 0 < b 3
Let the roots be a, a 2 .
Thus a, a 2 = 1 fi a 3 = 1
fi a = 1, w, w 2
When a = w,
p
w + w 2 = -
3
b
X
fi - 1 = -
p
3
fi p = 3
62. Given a, b are the roots of ax 2 + bx + c = 0.
b c
Thus, a + b = - , ab =
a a
Also, a + d, b + d are the roots of Ax 2 + Bx + C = 0.
B
Thus, a + d + b + d = -
A
C
and ( a + d) ◊ ( b + d)
= -
A
Now, a – b = (a + d) – (b + d)
fi (a – b) 2 = ((a + d) – (b + d)) 2
fi (a + b) 2 – 4ab = ((a + d) + (b + d)) 2 – 4(a + d)(b + d)
2 2
fi
b - 4 c = B -4
C
2 2
a a A A
2 2
b -4ac B -4AC
fi
=
2 2
a A
b c
63. Given a + b = - , ab =
a a
It is also given that a 2 x 2 + abc x + c 2 = 0
fi
fi
fi
3
c
x 0
3 3
2 abc
x + + =
a a
3
c
x 0
2 3
2 bc
x + + =
a a
3
2 b c c
x + Ê Á ˆÊ ˆ x+ Ê ˆ = 0
Ë
˜Á
a¯Ë ˜ Á ˜
a¯ Ëa¯
fi x 2 – (a + b)abx + (ab) 3 = 0
fi (x – a 2 b)(x – ab 2 ) = 0
Hence, the roots of the given equations are
a 2 b, ab 2
64. Here, a + b = 1, ab = p
and g + d = 4, gd = q
Also, a, b, g, d are in G.P .
Thus, b = a, r, g = ar 2 , d = ar 3
Now,
fi
fi
g + d
=
a + b
4
1
2 3
ar
+ ar
a + ar
= 4
ar
2 (1 + r)
= 4
a(1 + r)
fi r 2 = 4
fi r = ±2
When r = 2, a + b = 1
fi a + 2a = 1