快遞百科

標題:

if -3 is a root of the equation x^2+(13+k)x+k=0...

發問:

• 問BIO 2006 CE MC Q.10 , Q.12@1@
• Honda Euro R CL7問題 20分@1@
• Questions about formulas

此文章來自奇摩知識+如有不便請留言告知

if -3 is a root of the equation x^2+(13+k)x+k=0, find a) the value of k b) the other root of the equation. Thanks :)

最佳解答:

1)x^2+(13+k)x+k=0 (-3)^2+-3(13+k)+k=0 -2k=30 k=-15 2)x^2+(13+-(15))x-15=0 x^2-2x-15=0 (x-5)(x+3)=0 x=5 The other root of the equation is 5

其他解答:

Let a be another root of the equation Sum of roots = a - 3 = -(13+k) Product of roots = k = -3a Solving these two equations, we have a - 3 = -13+3a which gives a=5 Putting this into k=-3a, we have k = -15|||||a) (-3)^2+(13+k)(-3)+k=0 k=-15 b) x^2-2x-15=0 x=5 or x=-3 So the other root is x=5