Page 70 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

x 2 3 6 4
2.63 Solve the equation − x + + =0.
4 2 x x 2

Solution:
x 2 3 6 4 24 16 4 4
2
2
− x + + =0 ⇔ x − 6x + + =0 ⇔ (x − ) − 6(x − ) + 8 = 0.
4 2 x x 2 x x 2 x x
4 2
Let x − = y , then y − 6y +8 = 0 ⇔ (y − 2)(y − 4) = 0 ⇒ y =2 or y =4.
x
√
4 2
When y =2, we have x − =2 ⇒ x − 2x − 4= 0 ⇒ x =1 ± 5.
x
√
2
4 =4 ⇒ x − 4x − 4= 0 ⇒ x =2 ± 2 2.
When y =4, we have x −
x
√ √
Therefore, the original equation has four roots 1 ± 5, 2 ± 2 2.
2.64 m, n are positive integers, m = n , the equation
(m − 1)x − (m + 2)x +(m +2m)= 0 and the equation n m
2
2
2
(n − 1)x − (n + 2)x +(n +2n)=0 has a common root. Find the value of m + n .
2
2
2
m −n + n −m
Solution: The quadratic formula together with m> 1,n > 1,m = n gives us the
following: the first equation has roots x = m, m+2 , and the second equation has roots x
m−1
= n, n+2 . Since m = n , then m, n = n+2 ,n = m+2 . Both of these two equalities give us the
n−1 n−1 m−1
same result: mn − m − n − 2= 0 ⇔ (m − 1)(n − 1) = 3.

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