Finding integer a, b such that the roots of3x2+3(a+b)x+4ab=0satisfyα(α+1)+β(β+1)=(α+1)(β+1)

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Terzago66cl · Accepted Answer

Step 1Start fromα(α+1)+β(β+1)=(α+1)(β+1)Expand the parenthesis, and move all terms to one side:α2+α+β2+β−αβ−α−β−1=0I am going to write this in terms of the sum of the roots, α+β, and the product.α2+αβ+β2−1=0α2+2αβ+β2−αβ−1=0(α+β)2−αβ−1=0From Vieta&#039;s formula,α+β=−3(a+b)3=−(a+b)andαβ=4ab3From here[−(a+b)]2−4ab3−1=0a2+2b3a+(b2−1)=0a1,2=−b3±b29−(b2−1)=−b3±1−8b29For a to be real, b=0 or b=1. If b=0 then a=±1. Since you are told a&amp;gt;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 2As mentioned in the comments, there is a sign mistake in the sign after the second equation. The third equation should beα2+β2−αβ−1=0After that(α+β)2−3αβ−1=0With the Vieta,[−(a+b)]2−34ab3−1=0(a−b)2=1Since a&amp;gt;b, thena=b+1

Zaria Lamb · Accepted Answer

Step 1Because α and β are roots of 3x2+3(a+b)x+4ab, Vieta saysα+β=−a−b 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⇔(a−b)2=1 (3)Since a&amp;gt;b, we geta=b+1 (4)

Finding integer a, b such that the roots

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