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.

Question

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 $x^2-6x+11=0$

The answer in the book is what I find. – Bernard Nov 16 '15 at 12:42 — Bernard, Nov 16 '15 at 12:42

score 2 · Accepted Answer · answered Nov 16 '15 at 12:39

2

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

answered Nov 16 '15 at 12:39

mathlove

139,939

Woops. Now I feel silly. Thanks for pointing that out. – shredalert Nov 16 '15 at 12:42

score 1 · Answer 2 · answered Nov 16 '15 at 12:49

1

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.

answered Nov 16 '15 at 12:49

Mark Bennet

100,194

Will make note of the alternate approach under the conditions you specified. Thank you. – shredalert Nov 16 '15 at 12:53

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.

2 Answers2

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