? 發問於 科學及數學數學 · 1 十年前

1. Suppose a and b are two distinct roots of the equation x^2 +3x -2=0

(a) Find the values of a^2 +3a and b^2 +3b

*(b) Hence fin the value of a+b Ans:-3

2. Solve x^2+2ax+b=0

Ans: -a+開方(a^2-b) 或 -a-開方(a^2-b)

### 2 個解答

• cipker
Lv 5
1 十年前
最愛解答

1(a)

As a and b are the roots of x²+3x−2=0,

we have

a²+3a−2=0

b²+3b−2=0

a²+3a=2

b²+3b=2

(b)

Using the RESULT of(a),

a²+3a　　　　= 2

−) b²+3b　　　　= 2

-------------------------------

a² +3a−(b²+3b) =2−2

a²−b² +3a−3b=0

(a−b)(a+b)+3(a−b)=0

(a−b) [(a+b)+3]=0

a−b=0　　　　　or　(a+b)+3=0

a=b(rejected)　 or　　　a+b=−3

2

x²+2ax +b=0

x²+2ax+a²+b−a²=0

(x+a)² +b−a²=0

(x+a)² = a²−b

x+a = ±√(a²−b)

x=−a ±√(a²−b)

• Danny
Lv 6
1 十年前

2) x^2 + 2ax + b = 0

x^2 + 2ax + a^2 -a^2 + b = 0

( x + a )^2 - a^2 + b = 0

( x + a )^2 = a^2 - b

x + a = (a^2 - b)^(0.5) or x + a = - (a^2 - b)^(0.5)

x = - a + (a^2 - b)^(0.5) or - a - (a^2 - b)^(0.5)

1a) a is root of x^2 +3x -2=0: a^2 + 3a - 2 = 0 => a^2 + 3a = 2

b is root of x^2 +3x -2=0: b^2 + 3b - 2 = 0 => b^2 + 3b = 2

1b) By formula, sum of roots = a+b = -3/1 = - 3

Alternatively, using the result of Question 2,

sum of roots = [ - a + (a^2 - b)^(0.5) ] + [- a - (a^2 - b)^(0.5)]

= - 2 a

Compare x^2+2ax+b=0 and x^2 +3x -2=0

we get, a = 1.5

So sum of roots = -2a = -2*1.5 = -3