Page 56 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

√ √ √
2
2.21 Given y − ab = a bx − a + b a − bx (a> 0,b > 0), show log (xy ) =2.
a
Proof: The equation makes sense if and only if bx − a ≥ 0,a − bx ≥ 0, i.e. x ≥ a/b, x ≤ a/b ,
√
then x = a/b , substitute it into the original equation to obtain y = ab . Hence,
√
a
2
2
log (xy ) = log [ ( ab) ] = log ( · ab) = log a =2.
a
2
a
a
a b
a b
2.22 For what values of k, the quadratic equation (k − 1)x − 6(3k − 1)x + 72 = 0 with
2
2
variable x has two distinct positive integer roots.
Solution: Δ = 36(3k − 1) − 4 × 72(k − 1) = 36(k − 3) > 0 ⇒ k �=3 . The quadratic
2
2
2
6
12
formula implies x= 6(3k−1)±6(k−3) , that is, x1 = k+1 ,x 2 = k−1 . Since x 1 ,x 2 are positive
2
2(k −1)
integer roots and k =3, then k =2. When k =2, x 1 =4,x 2 =6. Hence, k =2 is the
only value of k such that the equation has two distinct positive integer roots.
4
5
2
2
2.23 Solve the equation P · C x+3 =(C − 1)P x+1 .
4
8
Solution:
4
5
2
2
P · C x+3 =(C − 1)P x+1
8
4
⇔

(x + 3)(x + 2)(x + 1)x 8 × 7 × 6 × 5 × 4
4 × 3 × = − 1 (x + 1)x
4 × 3 × 2 × 1 5 × 4 × 3 × 2 × 1
⇔
(x + 3)(x + 2)(x + 1)x
= 55(x + 1)x.
2
Since x +1 ≥ 2,x +3 ≥ 4, then x ≥ 1, then x �=0,x �= −1. We can divide (x + 1)x/2
2
on both sides: (x + 3)(x + 2) = 110 ⇔ x +5x − 104 = 0 ⇔ (x + 13)(x − 8) = 0 which
leads to x =8 or x = −13 (deleted). Hence, the original equation has the root x =8.

2.24 The three roots of the equation 3x + px + qx − 4 =0 are the side length, the
3
2
radius of the inscribed circle, the radius of the circumcircle, of a same equilateral triangle.
Find the values of p, q .

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