Page 66 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.52 If x,y,z are real roots of the equation system

2
x − yz − 8x +7 = 0,
2
2
y + z + yz − 6x +6 = 0,
find the range of x .

Solution: The system is equivalent to

2
yz = x − 8x + 7 (i)
2
2
y + z + yz =6x − 6 (ii)
2 2 2 2
2
2
(ii)-(i)×3: y +z −2yz = −3x +30x−27 ⇒ (y−z) = −3(x−1)(x−9) ≥ 0 ⇒ (x−1)(x−9) ≤
2
2
y +z −2yz = −3x +30x−27 ⇒ (y−z) = −3(x−1)(x−9) ≥ 0 ⇒ (x−1)(x−9) ≤
0 ⇒ 1 ≤ x ≤ 9
0 ⇒ 1 ≤ x ≤ 9 .
2.53 If a, b, k are rational numbers, and b = ak + , show the equation ax + bx + c =0
2
c
k
has two rational roots.
√
2
c 2
c 2
Proof: The discriminant Δ= b − 4ac =(ak + ) − 4ac =(ak − ), thus Δ= ±(ak − ).
c
k
k
k
c = ak − (b − ak)=2ak − b . Since a, b, k are rational numbers, ak −
c
In addition, ak − k k
√
is also a rational number, thus Δ is a rational number, therefore the two roots of the
√
quadratic equation x = −b± Δ are rational numbers.
2a
2 1 =0, find the
2.54 If x 1 ,x 2 are the two real roots of the equation x + ax + a − 2
value of a such that (x 1 − 3x 2 )(x 2 − 3x 1 ) reaches the maximum value.
Solution: Vieta’s formulas imply x 1 + x 2 = −a, x 1 x 2 = a − , thus
1
2
2
2
2
2
(x 1 −3x 2 )(x 2 −3x 1 ) = 10x 1 x 2 −3(x +x )=16x 1 x 2 −3(x 1 +x 2 ) = 16a−8−3a = 2
2
2
2
1 3(x +x )=16x 1 x 2 −3(x 1 +x 2 ) = 16a−8−3a =
2
(x 1 −3x 2 )(x 2 −3x 1 ) = 10x 1 x 2 −
2
1
8 2
−3(a − ) + 8 2 40 40 . Since the quadratic equation has two real roots, the discriminant
−3(a
3 − ) +
3
3
3
√
√
√
1
2
Δ= a − 4(a − ) = [(a − 2) + 2][(a − 2) − 2] ≥ 0 which leads to a ≥ 2+ 2 or
√ 2 √ √
a ≤ 2 − 2. Since 8/3 ∈ (2 − 2, 2+ 2), the extreme values should be obtained at
boundaries.
√ √
When a = 2+ 2, (x 1 − 3x 2 )(x 2 − 3x 1 )= 4 2+6.
√ √
When a =2 − 2, (x 1 − 3x 2 )(x 2 − 3x 1 )= −4 2+6.
√ √
Hence, when a = 2+ 2, (x 1 − 3x 2 )(x 2 − 3x 1 ) reaches the maximum value 4 2+6.
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