求quadratic equation 一解

If x² - ax + 5 ≡ (x - b)(x - c), where a, b and c are natural numbers. Evaluate a+b+c?

x² - ax + 5 ≡ (x - b)(x - c)

x² - ax + 5 ≡ x² - (b+c)x + bc

Comparing coefficients ,

a = b + c ..... (1)
and
bc = 5 .... (2)

By (2) , b = 1 , c = 5 or b = 5 , c = 1 , anyway , b + c = 6

By (1) , a + b + c = 2(b + c) = 2 * 6 = 12

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9a² - b² = 0 and ab<0 , then (a-b)/(a+b) = ?

(3a)² - b² = 0

(3a - b) (3a + b) = 0

3a - b = 0 or 3a + b = 0

3a = b (rejected since ab < 0 then a and b have difference sign)
or
3a = - b

then
(a - b) / (a + b)

= (a + 3a) / (a - 3a)

= 4a / (- 2a) ........... ab < 0 so a ≠ 0

= 4 / - 2

= - 2

x^2-ax+5 ≡ (x-b)(x-c)x^2-ax+5 ≡ x^2-(b+c)x+bc [ sum of roots and product of roots ]

comparing coefficients of x and constant,
b+c=a
bc=5

as a,b,c are natural numbers ( positive integers )
{b=5 and c=1} and {b=1 and c=5} are only possible solutions to bc=5
therefore, b+c=6
a+b+c=(b+c)+b+c=6+6=12

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9a^2-b^2=0
Method 1:

9a^2-b^2=0
(3a)^2-(b)^2=0
(3a)^2=b^2
3a=±b
Reject 3a=b since a*b=a*(3a)=3a^2>0

3a=-b
b=-3a
(a-b)/(a+b)=(a+3a)/(a-3a)=(1+3)/(1-3)=4/-2=-2
( "a" can be cancel out because ab<0, a≠0 )


2012-01-06 22:29:59 補充：
sorry for typing
no other method easier than this

Simon YAU the ans should be -2 rather than -1/2 ..

The second one is HKCEE Math past paper around 2001.
Simon YAU
http://hk-math.com