Page 59 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.30 Find all prime number solutions of the equation x(x + y)= z + 120.

Solution: When z =2, then x(x + y) = 122, then x + y = 122/x is an integer and since
x is a prime number, then x =2 or 61. When x =2, then y = 59; When x = 61, then

y = −59 (deleted).

When z is an odd number, then x and x + y are both odd numbers. Thus y has to be the

only even prime number, i.e. y =2. Then x(x + 2) = z + 120 ⇔ z =(x − 10)(x + 12).
Since z is a prime number, then x − 10 = 1, then x = 11,z = 23.

As a conclusion, there are two possibilities: x =2,y = 59,z =2 or x = 11,y =2,z = 23.

√ √
2.31 Solve the equation 2x − 7x +1 − 2x − 9x +4 = 1 (i).
2
2
√ √
Solution: Multiply both sides by 2x − 7x +1+ 2x − 9x +4:
2
2
√ √ √
2
2
2
2x − 7x +1+ 2x − 9x +4 = 2x − 3 (ii). (i)+(ii): 2x − 7x +1 = x − 1, taking
2
square to obtain x − 5x =0 ⇒ x(x − 5) = 0 ⇒ x =0 or x =5. We can verify these two
possible solutions via the original equation (i): x =5 is indeed a root of (i), but x =0 is
a extraneous root generated by taking square.
2.32 Find positive integers m, n such that the quadratic equation 4x − 2mx + n =0
2
has two real roots both of which are between 0 and 1.
2
2
Solution: The equation has two real roots, thus Δ=4m − 16n ≥ 0 ⇒ n ≤ m /4. Since
both roots are between 0 and 1, then f(0) = n> 0,f(1) = 4 − 2m + n> 0, then n> 2m − 4
(m, n ∈ N ). Hence, 2m − 4 <n ≤ m /4, which implies a unique choice: m =2,n =1.
2

2.33 Solve the system of equations

(1 + y) x = 100 (i),
(y − 1) 2x
2
4
(y − 2y + 1) x−1 = (y> 1) (ii).
(y + 1) 2
2x
Solution: (ii) ⇔ (y − 1) (y + 1) 2x = (y − 1) 2x . Since y �= ±1, then
2
(y − 1) (y + 1) 2 (y + 1) 2
(y−1) (y+1) −(y−1) (y−1) =0 ⇔ (y−1) [(y+1) −(y−1) ]=0 ⇒ y =1 or
2x
2
2x
2x
2
2x
2x
x
(y + 1) = ±(y − 1). The second case together with (i) leads to
±(y − 1) = 100 ⇒ y = 101 or y = −99 (deleted since y> 1).
2
x
When y =1, then (i) implies 2 = 100 ⇒ x = .
lg 2

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