1.Algebra Booster

2.54 Algebra Booster
n - 3 1
When <
2
n - 4 10
fi
2
fi
fi
fi
fi
n - 3 1
- < 0
n - 4 10
2
10n
- 30 – n + 4
0
2 <
10( n - 4)
10n
- 26 – n
2
( n - 4)
2
2
< 0
n - 10n+
26
> 0
2
( n - 4)
2
n - 10n+
26
> 0
( n+ 2)( n -2)
1
fi
> 0 , since D is negative.
( n+ 2)( n -2)
By the sign scheme, we get
n Œ (– , –2) » (2, )
From Eqs (i), (ii) and (iii), we get
…(iii)
3 29
3 < n < +
2 2
Hence, n = 4 (since n is a natural number) is the required
solution.
23. We have,
(15 4 14) x
x
+ + (15 - 4 14) = 30
t 1
fi (15 + 4 14) + = 30
t
(15 + 4 14)
1
fi a + = 30, where a = 15 + 4 14
a
Also, a 2 – 30a + 1 = 0
fi
30 ± 900 -4
a =
2
30 ± 8 14
= = 15 ± 4 14
2
When a = 15 + 4 14
fi
t
(15 + 4 14) = (15 + 4 14)
fi t = 1
fi x 2 – 2|x| = 1
fi |x| 2 – 2|x| – 1 = 0
fi
2± 4+
4
| x| = = 1±
2
2
fi x =± (1 ± 2)
When a = (15 -4 14)
t
fi (15 + 4 14) = (15 -4 14)
fi
t -1
(15 + 4 14) = (15 + 4 14)
fi t = –1
fi x 2 – 2|x| = –1
fi |x| 2 – 2|x| + 1 = 0
fi (|x| – 1) 2 = 0
fi (|x| – 1) = 0
fi |x| = 1
fi x = ±1
Hence the solution set is { ± (1 ± 2), -1, 1}.
24. Given x x+y = y n …(i)
y x+y = x 2n y n
…(ii)
Multiplying Eqs (i) and (ii), we get
(xy) x+y = (xy) 2n
fi x + y = 2n, when xy π 1
From Eqs (i) and (iii), we get
fi (x) 2n = y n
fi x 2 = y
From Eqs (iii), we get
x + x 2 = 2n
fi x 2 + x – 2n = 0
- 1± 1+
8n
fi x =
2
- 1+ 1+
8n
fi x =
2
2 1+ 1+ 8n
- 2 1+
8n
and y = x =
4
1
= (1 + 4 n - 1 + 8 n )
2
25. Given equations are
xy + 3y 2 – x + 4y – 7 = 0
…(i)
2xy + y 2 – 2x – 2y + 1 = 0
…(ii)
Multiplying Eq. (i) by 2 and subtract it from Eq. (ii), we get
–5y 2 – 10y + 15 = 0
fi y 2 + 2y – 3 = 0
fi (y + 3)(y – 1) = 0
fi y = –3, 1
When y = –3 , then –3x + 27 – x – 12 – 7 = 0
fi –4x + 8 = 0
fi x = 2
Hence the solutions are x = 2, y = –3; y = 1, x Œ R
26. We have,
a + b = p and ab = q
Now, sum of the roots
= (a 2 – b 2 )(a 3 – b 3 ) + a 3 b 3 + a 2 b 3
= (a – b) 2 (a + b)(a 2 + ab + b 2 ) + (ab) 2 (a + a)
= {(a + a) 2 – 4ab}p(p 2 – q) + q 2 p
= (p 2 – 4q)p(p 2 – q) + q 2 p
= p((p 2 – 4q)(p 2 – q) + q 2 )
= p(p 4 – 5p 2 q + 4q 2 )
and product of the roots
= (a 2 – b 2 )(a 3 – b 3 )(ab) 2 (a + a)