a maths question
2007-12-06 5:48 am
Given that a and b are the roots of a quadratic equation. If (a+2b)(b+2a)= -4 and a^2+b^2=13, find the quadratic equation.
回答 (2)
2007-12-06 6:01 am
✔ 最佳答案
(a+2b)(b+2a)= -4=> ab+2b^2+2a^2+4ab = -4
=> 2(a^2+b^2) + 5ab = -4
=> 26 + 5ab = -4
=> 5ab = -30
=> ab = -6
(a+b)^2 = a^2+2ab+b^2 = 13 + 2(-6) = 1
a+b = 1 or -1
so the equation is x^2 + x -6 = 0 or x^2 - x -6 = 0
參考: 膠力量
2007-12-06 6:15 am
(a+2b)(b+2a)= -4~~~ &~~ a^2+b^2=13
2a^2+5ab+2b^2= -4
2( a^2+b^2)+5ab= -4
2(13)+5ab= -4
+5ab= -30
ab = -6
ab = -6
a= -6/b ~~~ &~~~a^2+b^2=13
~~~~~~~~~~~~~~ 36/b^2+b^2=13
~~~~~~~~~~~~~~~~ 36+b^4=13b^2
~~~~~~~~~~~~~~~~ b^4 - 13b^2+36=0
~~~~~~~~~~~~~~~~ ( b^2 -4)( b^2 -9)=0
~~~~~~~~~~~~~~~~ b^2= 4 or 9
~~~~~~~~~~~~~~~~~ b = 2 or 3
(b=2), ~~~~~~~~~&~~(b=3),
a= -6/b~~~~~~~~~~~~ a= -6/b
a= -6/2 ~~~~~~~~~~~~a= -6/3
a= -3~~~~~~~~~~~~~~a = -2
so, (a,b)=(-3,2) or (a,b)=(-2,3)
2a^2+5ab+2b^2= -4
2( a^2+b^2)+5ab= -4
2(13)+5ab= -4
+5ab= -30
ab = -6
ab = -6
a= -6/b ~~~ &~~~a^2+b^2=13
~~~~~~~~~~~~~~ 36/b^2+b^2=13
~~~~~~~~~~~~~~~~ 36+b^4=13b^2
~~~~~~~~~~~~~~~~ b^4 - 13b^2+36=0
~~~~~~~~~~~~~~~~ ( b^2 -4)( b^2 -9)=0
~~~~~~~~~~~~~~~~ b^2= 4 or 9
~~~~~~~~~~~~~~~~~ b = 2 or 3
(b=2), ~~~~~~~~~&~~(b=3),
a= -6/b~~~~~~~~~~~~ a= -6/b
a= -6/2 ~~~~~~~~~~~~a= -6/3
a= -3~~~~~~~~~~~~~~a = -2
so, (a,b)=(-3,2) or (a,b)=(-2,3)
參考: myself
收錄日期: 2021-04-23 22:07:20
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