A.maths (quadratic equation and quadratic function)
2007-08-29 10:27 pm
if the roots of x^2-px+q=0 are two consecutive integers,prove that p^2-4q-1=0
回答 (3)
2007-08-29 10:36 pm
✔ 最佳答案
The roots of x^2-px+q=0 are two consecutive integersLet the roots be α and (α+1)
sum of the roots:
-(-p)÷1 = α + (α+1)
p = 2α + 1
product of the roots:
α(α+1) = q÷1
q = α(α+1)
∴LHS
=p^2-4q-1
=(2α+1)^2 - 4α(α+1) - 1
=4α^2 + 4α + 1 - 4α^2 - 4α - 1
=0
=RHS
2007-08-29 11:24 pm
Let the two roots be a and a+1
Sum of toot=p=2a+1
Product of root=q=a(a+1)
∴p^2-4q-1=4a^2+4a+1-4a^2-4a-1=0
Sum of toot=p=2a+1
Product of root=q=a(a+1)
∴p^2-4q-1=4a^2+4a+1-4a^2-4a-1=0
2007-08-29 10:41 pm
Sum of root= -(-p)/1=p
Product of root = q
If the roots are two consecutive integers, then their difference must be 1
so let the root be a and b
a-b= Sqrt( (a+b)^2-4ab ) = Sqrt (p^2-4q)=1
Squaring both side
p^2-4q=1
so p^2-4q-1=0
Product of root = q
If the roots are two consecutive integers, then their difference must be 1
so let the root be a and b
a-b= Sqrt( (a+b)^2-4ab ) = Sqrt (p^2-4q)=1
Squaring both side
p^2-4q=1
so p^2-4q-1=0
收錄日期: 2021-04-23 22:29:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070829000051KK02690