Page 75 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Solution: x + y + z =3 ⇔ x + y =3 − z (i), x + y + z =3 ⇔ x + y =3 − z (ii).
2
2
2
2
2
2
) (iii). (i) -(ii):
xy =( x+y 2 x−y 2 3−z 2 x−y 2 2 xy = (3−z) 2 − 3−z 2 (iv). (iii)&(iv)
) =(
) − (
) − (
2 2 2 2 2 2
2
2
) − (
⇒ ( 3−z 2 x−y 2 (3−z) 2 − 3−z 2 ⇒ 3(z − 1) +(x − y) =0 ⇒ z =1,x = y .
) =
2 2 2 2
Substitute them into (i) to obtain x = y =1. Obviously x = y = z =1 satisfies
x + y + z =3. Hence, the original system has the solution x =1,y =1,z =1.
5
5
5
2.77 Solve the equation 4x + 12x − 47x + 12x +4 = 0.
4
3
2
Solution: Obviously x =0 is not a root, so we assume x =0, then we can divide both
1
2
2
2
sides by x : 4x + 12x − 47 + 12 + 4 2 =0, then 4(x + 1 2 ) + 12(x + ) − 47 = 0 (i). Let
x x x x
2
2
x + 1 = u , then x + 1 2 = u − 2.
x x
Substitute them into (i) to obtain 4(u − 2) + 12u − 47 = 0 ⇒ 4u + 12u − 55 = 0 ⇒ u =5/2
2
2
2
or u = −11/2. When u =5/2, x + 1 = 5 ⇒ 2x − 5x +2 = 0 ⇒ x =2 or x =1/2. When
x 2
√
2
u = −11/2, x + 1 = − 11 ⇒ 2x + 11x +2 = 0 ⇒ x = −11± 105 . Hence, the original equation
x 2 4
√ √
has four roots: x =2,x =1/2,x = −11+ 105 ,x = −11− 105 .
4 4
√ √
2.78 Solve the equation 3 10 − 2x + 3 2x − 1= 3.
√ √ √
3
3
3
Solution: Let 10 − 2x = a, 3 2x − 1= b , then a + b =3. 10 − 2x = a ⇒ a = 10 − 2x
√
(i). 2x − 1= b ⇒ b =2x − 1 (ii). (i)+(ii)⇒ a + b =9 ⇒ (a + b)(a − ab + b ) =9 ⇒ a − ab + b =3
2
2
3
3
2
3
2
3
2
2
2
(iii). Substitute a =3 − b into (iii): (3 − b) − (3 − b)b + b =3 ⇒ b − 3b +2 = 0 ⇒ b =1
or b =2.
When b =1, (ii)⇒ x =1.
When b =2, (ii)⇒ x =9/2.
We can verify that x =1,x =9/2 are indeed two roots.
3
2
2.79 The real coefficient equation x +2kx +9x +5k =0 has an imaginary root
√
whose modulus is 5, find the value of k and solve the equation.

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