1.Algebra Booster

Quadratic Equations and Expressions 2.71
42. Given a + b = –p, ab = q
a 4 + b 4 = r, a 4 a 4 = s
Let the roots of x 2 – 4qx +2q 2 – r = 0 be g, d.
Thus, g, d = (2q 2 – r)
= 2(a, b) 2 – (a 4 + b 4 )
= –((a) 4 + (b) 4 – 2a 2 b 2 )
= –(a 2 – b 2 ) 2 < 0
So the roots are real and of opposite sign.
43. Since a is a root of a 2 x 2 + bx + c = 0
So, a 2 a 2 + ba + c = 0
Also b is a root of a 2 x 2 – bx – 2c = 0, so a 2 b 2 – bb – 2c = 0
Let f(x) = a 2 x 2 + 2bx + 2c
Thus, f(a) =a 2 a 2 + 2ba + 2c
= a 2 a 2 – 2a 2 a 2 = –a 2 a 2 < 0
Also, f(b)=a 2 b 2 + 2bb + 2c
= a 2 b 2 – 2a 2 b 2 = 3a 2 b 2 < 0
Therefore, f(a)f(b) < 0
Thus f(x) has one root which lies in (a, b)
i.e. a < g < b.
46. Given (x – a)(x – b) – c = (x – a)(x – b)
fi (x – a)(x – b) = (x – a)(x – b) – c
fi (x – a)(x – b) + c = (x – a)(x – b)
Hence the roots of (x – a)(x – b) + c = 0 are a and b.
49. Given p, q, r ΠAP.
2q = p + r
Since roots are real, so
D ≥ 0
fi q 2 – 4pr ≥ 0
fi
Ê p + r
Á
ˆ - 4pr
0
Ë
˜
2 ¯
fi (p + r) 2 – 16pr ≥ 0
fi p 2 + r 2 – 14pr ≥ 0.
fi
fi
2
2
Ê pˆ Ê pˆ
Á - 14 + 1≥0
Ë
˜ Á ˜
r¯ Ë r¯
Ê
Á
Ë
p
r
2
ˆ
-7˜
≥ 49 –1=
48
¯
Ê p ˆ
fi Á -7 ≥ 48
Ë
˜
r ¯
Ê p ˆ
fi Á -7 ≥4 3
Ë
˜
r ¯
51. We have
a
–1 1
af(–1) < 0 and af(1) < 0
fi a(a – b + c) < 0 and a(a + b + c) < 0
a
fi
2 2 b
a
Ê Á1- b + cˆ < 0anda
Ê 1+ + cˆ
< 0
Ë
˜
a a¯ Á
Ë
˜
a a¯
Ê b cˆ Ê b cˆ
fi Á1- + < 0and 1+ + < 0
Ë
˜
a a¯ Á
Ë
˜
a a¯
Ê b cˆ
fi Á1+ + < 0
Ë a a
˜
¯
Hence, the result.
53. Clearly, it has no solution.
Y
X¢ X
O 1/4 1
Y¢
54. We have 2 |y| – |2 y–1 – 1| = 2 y–1 + 1
fi 2 |y| – |2 y–1 + 1| + |2 y–1 – 1|
fi |2 y–1 + 1| = |2 y–1 – 1| = 2 |y|
fi (2 y–1 + 1)(2 y–1 – 1) ≥ 0
fi (2 2(y–1) – 1 2 ) ≥ 0
fi (2 2(y–1) ) ≥ 2 0
fi y – 1 ≥ 0
fi y ≥ 1
When y < 0, the given equation reduces to
2 –y – (1 – 2 y–1 ) = 2 y–1 + 1
fi 2 –y – 1 + 2 y–1 = 2 y–1 + 1
fi 2 –y = 2
fi –y = 1
fi y = –1
Hence the solution set is x Œ [1, ) » {–1}.
55. Given equation is
|x – 2| 2 + |x – 2| – 2 = 0
fi a 2 + a – 2 = 0, a = |x – 2|
fi (a + 2)(a – 1) = 0
fi (a + 2) = 0, (a – 1) = 0
fi a = 1, a = –2
fi |x – 2| = 1 and |x – 2| = –2
Thus, |x – 2| = 1
fi x – 2 = ±1
fi x = 2 ± 1 = 3, 1
57. On solving we get,
1 2 5
u =- , v = , w=
3 3 3
It is given that a, b, c, d are in G.P
So, let b = ar, c = ar 2 , d = ar 3
We have Î(b – c) 2 + (c – a) 2 + (d – b) 2˚
= (ar – ar 2 ) 2 + (ar 2 – a) 2 + (ar 3 – ar) 2
= (a 2 r 2 – 2a 2 r 3 + a 2 r 4 ) + (a 2 r 4 – 2a 2 r 2 + a 2 )
+ (a 2 r 6 – 2a 2 r 4 + a 2 r 2 )
= a 2 (r 2 – 2r 3 + r 4 + r 4 – r 2 + 1 + r 6 – 2r 4 + r 2 )
= a 2 (r 6 – 2r 3 + 1)