Page 72 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Proof: The equation represents two straight lines, then we should have
2 2 2 2 .
x −y +dx+ey+f =(x−y+k 1 )(x+y+k 2 )= x −y +(k 1 +k 2 )x+(k 1 −k 2 )y+k 1 k 2
Make the corresponding coefficients equal: k 1 + k 2 = d, k 1 − k 2 = e, k 1 k 2 = f . The first
two equations lead to k 1 = d+e ,k 2 = d−e , and substitute them into the third equation:
2 2
2
2
d+e · d−e = f , which is equivalent to d − e − 4f =0.
2 2

2
2
3
2.69 Solve the equation log (x + 1) − log xy + log √ 2 y +4 = 3.
8
2
Solution:

2
2
3
2
2
log (x +1) −log xy +log √ 2 y +4 = 3 ⇔ log (x +1)−log xy +log (y +4) =
2
8
2
2
2
2
2
2
2
3 ⇔ log 2 (x +1)(y +4) =3 ⇔ (x +1)(y +4) =8 .
xy
xy
Since x, y =0, we have
x y +4x + y +4 = 8xy ⇔ (2x − y) +(xy − 2) =0 ⇒ 2x − y =0, xy − 2= 0.
2
2
2
2 2
2
Solve these two equations to obtain two solutions of the original equation:
(1, 2), (−1, −2).
√ √ √
2.70 Solve the equation x + x +7 +2 x +7x = 35 − 2x .
2
Solution:
√ √ √ √ √
2
x+ x + 7+2 x +7x = 35−2x ⇔ x+2 x(x + 7)+x+7+ x+ x +7−42 =
√
√
√
√
√
√
√
√
0 ⇔ ( x+ x + 7) +( x+ x + 7)−42 = 0 ⇔ ( x+ x + 7+7)( x+ x +7−6) = 0.
2
√ √ √ √
Since x + x +7 +7 > 0 , then x + x +7 = 6 . Squaring both sides to obtain
√
2 x +7x = 29 − 2x , and squaring again to obtain 144x = 841, thus x = 841/144 which
2
is the root of the original equation.
2.71 The x -dependent equation x + p|x| = qx − 1 has four distinct real roots,
2
show that p + |q| < −2.
2
Proof: When x> 0, the equation becomes x +(p − q)x +1 = 0 (i); When x< 0, the
equation becomes x − (p + q)x +1 = 0 (ii). We need two positive roots from (i) and two
2
negative roots from (ii). Hence, the smaller root of (i) is greater than zero, and the larger
√ √
2
2
root of (ii) is less than zero, that is q−p− (p−q) −4 > 0 and p+q+ (p+q) −4 < 0 (obviously
2 2
both discriminants need to be positive, (p − q) − 4 > 0, (p + q) − 4 > 0 ). Therefore,
2
2

2
2
q − p> (p − q) − 4 > 0 (iii) and 0 < (p + q) − 4 < −(p + q) (iv). (iii) implies
q >p , and since (p − q) − 4 > 0 , then q − p> 2 , then p − q< −2 . (iv) implies
2
p + q< 0, and since (p + q) − 4 > 0, then p + q< −2. As a conclusion, p + |q| < −2.
2
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