if k is a root of the quadratic equation 6x^2 +3kx -4=0
find the values of k
maths question
2007-09-25 4:51 pm
回答 (2)
2007-09-25 5:13 pm
✔ 最佳答案
As the question mentioned, we can put x=k into the equation such that...6(k)^2 + 3k(k) - 4 = 0
6k^2 + 3k^2 - 4 = 0
9k^2 - 4 = 0
(3k)^2 - (2)^2 = 0
(3k+2)(3k-2) = 0 ( Since (A^2-B^2 = (A+B)(A-B) )
k = -2/3 or k = 2/3
參考: Me
2007-09-25 5:12 pm
roots={-3k+[(3k)^2-4(6)(-4)]^1/2}/2(6)=k
-3k+[(3k)^2-4(6)(-4)]^1/2=12k
[(3k)^2-4(6)(-4)]^1/2=15k
[(3k)^2-4(6)(-4)]=15^2(k^2)=225k^2
9k^2+96=225k^2
96=216k^2, k^2=0.4444444, k=0.67 or -0.67
-3k+[(3k)^2-4(6)(-4)]^1/2=12k
[(3k)^2-4(6)(-4)]^1/2=15k
[(3k)^2-4(6)(-4)]=15^2(k^2)=225k^2
9k^2+96=225k^2
96=216k^2, k^2=0.4444444, k=0.67 or -0.67
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070925000051KK00674