標題:
求quadratic equation 一解
發問:
If x^2 ax 5 ≡ (x-b)(x-c), where a, b and c are natural numbers. Evaluate a+b+c? Plz help me with this one, steps should be clearly shown. thanks a lot 更新: Sorry, it should be: If x^2 - ax + 5 ≡ (x-b)(x-c) 更新 2: 還有一題有待指教 9a^2 - b^2 =0 and ab
最佳解答:
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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 ************************************************************************ 9a2 - b2 = 0 and ab
其他解答:
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 ======================================= 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
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