Page 88 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

x
x
Solution: Multiply the equation by 2a and reorganize it to obtain a 2x − 2ma +1 = 0. Let
t = a (t> 0 ), then t − 2mt +1 = 0 . When Δ=4m − 4 ≥ 0 , i.e. m ≥ 1 or
2
2
x
√ √
2
2
m ≤−1, the t - dependent equation has real roots: t 1 = m − m − 1,t 2 = m + m − 1.
If m =1, then t 1 = t 2 =1, thus a =1, the original equation has a unique root x =0. If
x
√ √ √
x
2
2
2
m> 1, m + m − 1 >m − m − 1 > 0, then a = m ± m − 1, that is, the original
√
2
equation has two distinct real roots: x = log (m ± m − 1) . If m< 1 ,
a
√ √ √ √
2
2
2
2
since |m| > m − 1, then m< − m − 1, then m − m − 1 ≤ m + m − 1 < 0 .
√
x 2 x
a = m ± m − 1 has no solution since a > 0. As a conclusion, when m< 1, the original
equation has no root; when m =1 , the original equation has a unique root x =0; when

√
2
m> 1, the original equation has two distinct roots x = log (m ± m − 1).
a
2.99 Solve the system of equations
4
2
x + y + z = 18
2
x y − yz 1/2 = −3
z 1/2 2 =4
x
to obtain real solutions.

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