Finding integer a, b such that the roots

Erika Bernard

Erika Bernard

Answered question

2022-04-05

Finding integer a, b such that the roots of
3x2+3(a+b)x+4ab=0
satisfy
α(α+1)+β(β+1)=(α+1)(β+1)

Answer & Explanation

Terzago66cl

Terzago66cl

Beginner2022-04-06Added 8 answers

Step 1
Start from
α(α+1)+β(β+1)=(α+1)(β+1)
Expand the parenthesis, and move all terms to one side:
α2+α+β2+βαβαβ1=0
I am going to write this in terms of the sum of the roots, α+β, and the product.
α2+αβ+β21=0
α2+2αβ+β2αβ1=0
(α+β)2αβ1=0
From Vieta's formula,
α+β=3(a+b)3=(a+b)
andαβ=4ab3
From here
[(a+b)]24ab31=0
a2+2b3a+(b21)=0
a1,2=b3±b29(b21)=b3±18b29
For a to be real, b=0 or b=1. If b=0 then a=±1. Since you are told a>b, it means only a=1 is good. If b=1, a=13±13. One solution is not integer, the other is less than b.
Step 2
As mentioned in the comments, there is a sign mistake in the sign after the second equation. The third equation should be
α2+β2αβ1=0
After that
(α+β)23αβ1=0
With the Vieta,
[(a+b)]234ab31=0
(ab)2=1
Since a>b, then
a=b+1
Zaria Lamb

Zaria Lamb

Beginner2022-04-07Added 8 answers

Step 1
Because α and β are roots of 3x2+3(a+b)x+4ab, Vieta says
α+β=ab and αβ=43ab (1)
Since α(α+1)+β(β+1)=(α+1)(β+1), we have
(α+β)2=3αβ+1 (2)
Plugging (1) into (2), we have
(a+b)2=4ab+1(ab)2=1 (3)
Since a>b, we get
a=b+1 (4)

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