Page 69 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.60 The equation 5x − (10 cos α)x + 7 cos α +6 = 0 has two identical roots, α is
2
one angle of a parallelogram, and the sum of two adjacent sides is 6, find the maximal area
of the parallelogram.

Solution: The quadratic equation has two identical roots, thus the discriminant
2
2
Δ = 100 cos α − 140 cos α − 120 = 0 ⇔ 5 cos α − 7 cos α − 6 =0 ⇒ cos α = 7±13 . Since
10
3
0
| cos α|≤ 1, then cos α = 7−13 = − . The angle of a parallelogram, α , is between 0 and
10 5
√
3
4
0
0
0
2
180 , and since cos α = − < 0, thus α ∈ (90 , 180 ), then sin α = 1 − cos α = . Let
5 5
one side of parallelogram has length u , then one adjacent side has length 6 − u . The area
4
4
2
S = u(6 − u) sin α = u(6 − u) = − (u − 3) + 36 . Hence, the maximal area S max = 36
5 5 5 5
when u =3.
2.61 Find all positive integer solutions (x, y) of the equation
√ √ √ √ √
x y + y x − 2011x − 2011y + 2011xy = 2011.
Solution: The equation is equivalent to √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ 2 2 √ √ √ √ √ √
√
xy( x+ y)−
xy x+ xy y− 2011x− 2011y+ 2011xy−( 2011) =0 ⇔ xy( x+ y)−
xy x+ xy y− 2011x− 2011y+ 2011xy−( 2011) =0 ⇔
√
√
√
√
√ √
√
√
√
√ √
√
√
√
√
√
√
√
√
√
2
√ √ √ √ √ √ √ √ √ √ 2 √ √ √ √ √ 2011( x+ y)+ 2011 xy−( 2011) =0 ⇔ ( xy− 2011)( x+ y+ 2011) = 0
2
2011( x+ y)+ 2011 xy−( 2011) =0 ⇔ ( xy− 2011)( x+ y+ 2011) = 0
√
√
√
√
√
√
√
√
xy x+ xy y− 2011x− 2011y+ 2011xy−( 2011) =0 ⇔ xy( x+ y)− √
2
√
√
√ √ xy x+ xy y− 2011x− 2011y+ 2011xy−( 2011) =0 ⇔ √xy( x+ y)−
√
√
√
√
√
√
2
√
√
2011( x+ y)+ 2011 xy−( 2011) =0 ⇔ ( xy− 2011)( x+ y+ 2011) = 0
√
√
√
√
√
√
√
√
√
2011( x+ y)+ 2011 xy−( 2011) =0 ⇔ ( xy− 2011)( x+ y+ 2011) = 0.
2
√ √ √ √ √
Since x + y + 2011 > 0, then xy − 2011 = 0 ⇒ xy = 2011. Since 2011 is a
prime, then x =1,y = 2011 or x = 2011,y =1. Hence the original equation has two
positive integer solutions (1, 2011), (2011, 1).
2.62 x 1 ,x 2 are two roots of the quadratic equation x − (k − 2)x + k +3k +5 = 0
2
2
where k is a real number, find the maximum value of x + x .
2
2
1
2
Solution: According to Vieta’s formulas, we have x 1 + x 2 = k − 2,x 1 x 2 = k +3k +5, thus
2
x + x =(x 1 + x 2 ) − 2x 1 x 2 =(k − 2) − 2(k +3k + 5) = −(k + 5) + 19.
2
2
2
2
2
2
2
1
Since the equation has real roots, then the discriminant
2
2
2
Δ=(k − 2) − 4(k +3k + 5) ≥ 0 ⇔ 3k + 16k + 16 ≤ 0 ⇒−4 ≤ k ≤− . The function
4
3
4
f(k) = −(k + 5) + 19 is a monotonically decreasing function on the interval [−4, − ],
2
3
thus the maximum value is f(−4) = 18 which is also the maximum value of x + x .
2
2
1
2
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