[ 數學 ]幫忙解數學題目

上面兩份解答都是錯的
先整理所求a^2+3ab+2b^2 = (a+b)^2+ab+b^2 = (a+b)[(a+b)+b], 所以我們必需求a+b和b
∵(a+b)^2=a^2+2ab+b^2, (a-b)^2=a^2-2ab+b^2 ∴(a+b)^2=(a-b)^2+4ab=97, a+b=√97或-√97, 所以我們必須從已知條件判斷, ∵a-b=7>0 ∴a>b; ∵ab=12>0 ∴a,b皆為正數或皆為負數, 兩個條件加總就是a,b皆為正數或皆為負數但a>b, 故a+b仍=√97或-√97, 此式與a-b=7聯立對應得b= (√97-7)/2 或 (7-√97)/2, 代回所求的簡化式即為所求：
(291-7√97)/2 或 (291+7√97)/2

已知a-b＝7及 ab＝12 求 a^2+3ab+2b^2=


(a+b)^2=(a-b)^2+4ab=49+48=93
ab=12 --> a, b 同正或負號

當 a+b=sqrt(93) ...(1)
a-b=7 ...(2)
b=[sqrt(93)-7]/2 --> a+2b=[3sqrt(93)-7]/2

a^2+3ab+2b^2
=(a+b)(a+2b)
=sqrt(93)[3sqrt(93)-7]/2
=[279-7sqrt(93)]/2

當 a+b= - sqrt(93) ...(3)
(3)-(2)
b=[ - sqrt(93) - 7]/2 --> a+2b= [- 3sqrt(93) - 7]/2

a^2+3ab+2b^2
=(a+b)(a+2b)
= - sqrt(93)[ - 3sqrt(93)-7]/2
=[279+7sqrt(93)]/2

Ans: [279-7sqrt(93)]/2 or [279+7sqrt(93)]/2

a > b a + (-b) = 7, a(-b) = -12 a, -b 為 x^2 – 7x -12 = 0之兩根 x = (7±√97)/2 a = (7+√97)/2, -b = (7-√97)/2, b = (-7+√97)/2 a^2 + 3ab + 2b^2 = (a + b)(a + 2b) = (√97)(-7+3√97)/2 = (291 - 7√97)2