1.Algebra Booster

4.74 Algebra Booster
2
Ê a ˆ =
2 2
a m +
Á3am
+ 9 ( 1)
Ë
˜ m ¯
a
9am 6a 9am 9a
m
a
2
2 2 2 2 2 2
+ + = +
2
2
m
a
2
2
m
a
m
2
2
2
2 2
+ 6a
= 9a
2 2 2
= 9a - 6a = 3a
= 3a
2
2 1
m =
3
1
m =±
3
Thus, the equations of the common tangents are
1
y =± ( x- a ) ± a 3
3
3 3
24. Given zz + z z = 350
2 2
fi zz ( z + z ) = 350
fi 2(x 2 – y 2 )(x 2 + y 2 ) = 350
fi (x 2 – y 2 )(x 2 + y 2 ) = 175
fi (x 2 – y 2 )(x 2 + y 2 ) = 7 ¥ 25
fi (x 2 – y 2 ) = 7, (x 2 + y 2 ) = 25
Thus x = 4, y = 3
Hence, the area of a rectangle = (2x ◊ 2y)
= (4xy)
= 48 sq. u.
25. Given b n = 1
Let S n
= 1 + 3b + 5b 2 + … + (2n – 1)b n – 1
fi S n
b = b + 3b 2 + … + (2n – 3)b n – 1 + (2n – 1)b n
Subtracting, we get,
(1 – b)S n
= 1 + 2b + 2b 2 + … + 2b n – 1 – (2n – 1)b n
= 1 + 2b + 2b 2 + … + 2b n – 1 – (2n – 1)
= 2b + 2b 2 + … + 2b n – 1 – 2n
Ê
n
1 - b ˆ
= 2Á
- 2n
Ë 1 - b
˜
¯
= –2n ( b n = 1)
fi
2n
Sn
=
b - 1
26. Given x 3 = –9 + 46i
fi (a + ib) 3 = –9 + 46i
fi a 3 + i3a 2 b – 3ab 2 – ib 3 = –9 + 46i
fi (a 3 – 3ab 2 ) + i(3a 2 b – b 3 ) = –9 + 46i
fi (a 3 – 3ab 2 ) = –9, (3a 2 b – b 3 ) = 46
fi a(a 2 – 3b 2 ) = –9, b(3a 2 – b 2 ) = 46
fi a = 3, b = 2
Hence, the value of
(a 3 + b 3 ) = 27 + 8 = 35.
27. We have w 5 = 2
Now, x = w + w 2
fi x 5 = (w + w 2 ) 5
fi x 5 = w 5 + 5w 6 + 10w 7 + 10w 8 + 5w 9 + w 10
fi x 5 = 2 + 10w + 20w 2 + 20w 3 + 10w 4 + 4
fi x 5 = 6 + 10(w 2 + 2w 3 + w 4 ) + 10(w 2 + w)
fi x 5 = 6 + 10(w + w 2 ) 2 + 10(w 2 + w)
fi x 5 = 6 + 10x 2 + 10x
fi x 5 – 10x 2 – 10x = 6.
28. Given equation is
z 2 – (3 + i)z + m + 2i = 0
It will provide us the real solution only when
z 2 – 3z + m = 0 and 2 – z = 0
fi z 2 – 3z + m = 0 and z = 2
Thus, m = 3z – z 2
fi m = 3 ¥ 2 – 4 = 2
Hence, the value of m is 2.
29. We have P(2) = 0
fi 32 + 8a + 4b + 2c + 3 = 0
fi 8a + 4b + 2c = –35 ...(i)
Also, P(i) = 2i 4 + ai 3 + bi 2 + ci + 3 = 0
fi 2 – ai – b + ci + 3 = 0
fi (5 – b) + i(c – a) = 0
fi (5 – b) = 0, (c – a) = 0
fi b = 5, c = a ...(ii)
From Relations (i) and (ii), we get
8a + 4b + 2c = –35
fi 8a + 20 + 2a = –35
fi 10a = –55
fi
55 11
a =- =-
10 2
30. We have z 5 + 1 = 0
Hence, the solutions of z are
± ip
3
{ 1, }
5 ± i p
- e , e
5
= {- 1, aa , , b, b}
, where a, w, ΠC
Êp
ˆ
Now, a + a = 2cos Á , a ◊ a = 1
Ë
˜
5 ¯
Ê3ð
ˆ
and b+ b= 2cos Á , b◊ b=
1
Ë
˜
5 ¯
Thus, z 5 + 1
( )
= ( z + 1)( z -a) z -a ( z -b)( z -b)
2
= ( z + 1)( z -( a + a) z + a◊a)
2
[ z -( b + b) z + b ◊b]
È 2 Êp
ˆ ˘
= ( z + 1) Íz - 2cosÁ
˜ z + 1
5
˙
Î Ë ¯ ˚
È 2 Ê3p
ˆ ˘
Íz
- 2 cosÁ
˜ z + 1
Ë 5 ¯ ˙
Î
˚