The roots of $x^2-2x+3=0$ are $\alpha$ and $\beta$. Find the equation whose roots are: $\alpha+2$, $\beta+2$. Not sure of answer in book. [duplicate]

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The roots of $x^2-2x+3=0$ are $\alpha$ and $\beta$. Find the equation whose roots are: $\alpha+2$, $\beta+2$. Not sure of answer in book.

My working:

$\alpha+\beta=2, \alpha\beta=3$

$(\alpha+2)+(\beta+2)=\alpha+\beta+4=6$

$(\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)=7$

Therefore, equation whose roots are $\alpha+2, \beta+2$ is $x^2-6x+7=0$

Answer in book is $x^2-6x+11=0$

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2 Answers

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Note that $$(\alpha+2)(\beta +2)=\alpha\beta+2(\alpha+\beta)\color{red}{+4}$$

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Another approach.

Let $y=x+2$ so that $x=y-2$ and $(y-2)^2-2(y-2)+3=0$ or $y^2-6y+11=0$

Such an approach may be possible when all the roots of the first equation are transformed in the same way to get the second, and sometimes has advantages of simplicity and of doing all the coefficients at once.

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