Question

已知a^2+b^2=12，a+b=4，求a、b

初一数学-平方差或完全平方公式算
3分钟后下线

Answer 1

a+b=4
两边同时平方，得
a²+2ab+b²=16
12+2ab=16
ab=2
所以
a,b是方程
x²-4x+2=0
的两个不同的实数根
所以
a,b=(4±√8)/2=2±√2

Answer 2

(a+b)^2 = 16
a^2+b^2 +2ab =16
2ab=4
ab = 2
a+ 2/a =4
a^2-4a+2=0
a= 2+√2 or 2- √2
when a= 2+√2, b =2-√2
when a= 2-√2, b =2+√2

Answer 3

a+ b = 4
(a+b)^2 = 16
a^2+b^2 = 12
2ab = 4
ab = 2

a(4-a) =2
4a - a^2 = 2
a^2 - 4a + 2 = 0
a^2 - 4a + 4 = 2
a -2 = ±√2
a = 2 ±√2

(1) a = 2 +√2 ，b = 2 - √2
(2) a = 2 -√2 ，b = 2 + √2

Answer 4

解：a＋b＝4
﹙a＋b﹚²＝4²
a²＋2ab＋b²＝16
12＋2ab＝16
2ab＝4
a·b＝2