Erika Bernard

2022-04-05

Finding integer a, b such that the roots of

$3{x}^{2}+3(a+b)x+4ab=0$

satisfy

$\alpha (\alpha +1)+\beta (\beta +1)=(\alpha +1)(\beta +1)$

satisfy

Terzago66cl

Beginner2022-04-06Added 8 answers

Step 1

Start from

$\alpha (\alpha +1)+\beta (\beta +1)=(\alpha +1)(\beta +1)$

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.

${\alpha}^{2}+\alpha \beta +{\beta}^{2}-1=0$

${\alpha}^{2}+2\alpha \beta +{\beta}^{2}-\alpha \beta -1=0$

${(\alpha +\beta )}^{2}-\alpha \beta -1=0$

From Vieta's formula,

$\alpha +\beta =-\frac{3(a+b)}{3}=-(a+b)$

and$\alpha \beta =\frac{4ab}{3}$

From here

${[-(a+b)]}^{2}-\frac{4ab}{3}-1=0$

${a}^{2}+\frac{2b}{3}a+({b}^{2}-1)=0$

$a}_{1,2}=-\frac{b}{3}\pm \sqrt{\frac{{b}^{2}}{9}-({b}^{2}-1)}=-\frac{b}{3}\pm \sqrt{1-\frac{8{b}^{2}}{9}$

For a to be real,$b=0$ or $b=1$ . If $b=0$ then $a=\pm 1$ . Since you are told $a>b$ , it means only $a=1$ is good. If $b=1,\text{}a=-\frac{1}{3}\pm \frac{1}{3}$ . 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

${(\alpha +\beta )}^{2}-3\alpha \beta -1=0$

With the Vieta,

${[-(a+b)]}^{2}-3\frac{4ab}{3}-1=0$

${(a-b)}^{2}=1$

Since$a>b$ , then

$a=b+1$

Start from

Expand the parenthesis, and move all terms to one side:

I am going to write this in terms of the sum of the roots,

From Vieta's formula,

and

From here

For a to be real,

Step 2

As mentioned in the comments, there is a sign mistake in the sign after the second equation. The third equation should be

After that

With the Vieta,

Since

Zaria Lamb

Beginner2022-04-07Added 8 answers

Step 1

Because$\alpha$ and $\beta$ are roots of $3{x}^{2}+3(a+b)x+4ab$ , Vieta says

$\alpha +\beta =-a-b$ and $\alpha \beta =\frac{43}{a}b$ (1)

Since$\alpha (\alpha +1)+\beta (\beta +1)=(\alpha +1)(\beta +1)$ , we have

${(\alpha +\beta )}^{2}=3\alpha \beta +1$ (2)

Plugging (1) into (2), we have

${(a+b)}^{2}=4ab+1\iff {(a-b)}^{2}=1$ (3)

Since$a>b$ , we get

$a=b+1$ (4)

Because

Since

Plugging (1) into (2), we have

Since

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