Erika Bernard

2022-04-05

Finding integer a, b such that the roots of
$3{x}^{2}+3\left(a+b\right)x+4ab=0$
satisfy
$\alpha \left(\alpha +1\right)+\beta \left(\beta +1\right)=\left(\alpha +1\right)\left(\beta +1\right)$

Terzago66cl

Step 1
Start from
$\alpha \left(\alpha +1\right)+\beta \left(\beta +1\right)=\left(\alpha +1\right)\left(\beta +1\right)$
Expand the parenthesis, and move all terms to one side:
${\alpha }^{2}+\alpha +{\beta }^{2}+\beta -\alpha \beta -\alpha -\beta -1=0$
I am going to write this in terms of the sum of the roots, $\alpha +\beta$, and the product.
${\alpha }^{2}+\alpha \beta +{\beta }^{2}-1=0$
${\alpha }^{2}+2\alpha \beta +{\beta }^{2}-\alpha \beta -1=0$
${\left(\alpha +\beta \right)}^{2}-\alpha \beta -1=0$
From Vieta's formula,
$\alpha +\beta =-\frac{3\left(a+b\right)}{3}=-\left(a+b\right)$
and$\alpha \beta =\frac{4ab}{3}$
From here
${\left[-\left(a+b\right)\right]}^{2}-\frac{4ab}{3}-1=0$
${a}^{2}+\frac{2b}{3}a+\left({b}^{2}-1\right)=0$
${a}_{1,2}=-\frac{b}{3}±\sqrt{\frac{{b}^{2}}{9}-\left({b}^{2}-1\right)}=-\frac{b}{3}±\sqrt{1-\frac{8{b}^{2}}{9}}$
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 . 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
${\alpha }^{2}+{\beta }^{2}-\alpha \beta -1=0$
After that
${\left(\alpha +\beta \right)}^{2}-3\alpha \beta -1=0$
With the Vieta,
${\left[-\left(a+b\right)\right]}^{2}-3\frac{4ab}{3}-1=0$
${\left(a-b\right)}^{2}=1$
Since $a>b$, then
$a=b+1$

Zaria Lamb

Step 1
Because $\alpha$ and $\beta$ are roots of $3{x}^{2}+3\left(a+b\right)x+4ab$, Vieta says
$\alpha +\beta =-a-b$ and $\alpha \beta =\frac{43}{a}b$ (1)
Since $\alpha \left(\alpha +1\right)+\beta \left(\beta +1\right)=\left(\alpha +1\right)\left(\beta +1\right)$, we have
${\left(\alpha +\beta \right)}^{2}=3\alpha \beta +1$ (2)
Plugging (1) into (2), we have
${\left(a+b\right)}^{2}=4ab+1⇔{\left(a-b\right)}^{2}=1$ (3)
Since $a>b$, we get
$a=b+1$ (4)

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