The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$.

Question

Quadratic equation question, as specified in the title.

The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.

I gather that $α + β = -\frac{b}{a} = \frac{6}{1} = 6$ and that $αβ = \frac{c}{a} = \frac{7}{1} = 7$. Do I need to convert $α + \frac{1}{β}$ and $β + \frac{1}{α}$ into a format whereby I can sub in the values for adding together or multiplying $α$ and $β$ ? If so, how ?

Expand $(\alpha+1/\beta)+(\beta+1/\alpha)$ and $(\alpha+1/\beta)(\beta+1/\alpha)$. — Kenny Lau, Jun 23 '16 at 16:38

score 4 · Answer 1 · answered Jun 23 '16 at 16:40

4

As $\alpha\beta=\dfrac71,$

let $y=\dfrac{\alpha\beta+1}\alpha=\dfrac{7+1}\alpha\iff\alpha=\dfrac8y$

Put the value of $\alpha$ in $$x^2-6x+7=0$$ and rearrange.

answered Jun 23 '16 at 16:40

lab bhattacharjee

274,582

Thank you. I thought of something along these lines, but did not realize that 1 refers strictly to the number 1 in this context. – Mad Banners Jun 23 '16 at 16:43
@MadBanners, See http://www.iitjeemathematics.com/conte/quadratic-equations/14-transformation-of-equations – lab bhattacharjee Jun 23 '16 at 16:44
Would the answer to this one be x^2 - 8x + 4/3 = 0? – Mad Banners Jun 23 '16 at 16:49

score 3 · Accepted Answer · answered Jun 23 '16 at 16:49

3

If $\alpha,\beta$ are the roots of $x^2-6x+7=0$, then all quadratic equations with roots $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$ are $$a\left(x-\left(\alpha+\frac{1}{\beta}\right)\right)\left(x-\left(\beta+\frac{1}{\alpha}\right)\right)=0,$$

where $a\in\mathbb R$, $a\neq 0$ (saying "the equation" is wrong, because there are infinitely many of them).

Using Vieta's Formulas, we get $\alpha+\beta=6$ and $\alpha\beta=7$. Use these two equalities in the following:

$$a\left(x^2-\left(\alpha+\beta+\frac{\alpha+\beta}{\alpha\beta}\right)x+\left(\alpha\beta+1+1+\frac{1}{\alpha\beta}\right)\right)=0$$

answered Jun 23 '16 at 16:49

user236182

13,324

So, x^2 - 48/7x + 64/7 = 0? – Mad Banners Jun 23 '16 at 16:54
1

@MadBanners Right, that's one of the equations that the problem asks for. A more beautiful equation would be $7x^2-48x+64=0$. – user236182 Jun 23 '16 at 16:55
Ah, I see. Thank you very much! – Mad Banners Jun 23 '16 at 16:56
Why the downvote? – user236182 Jul 27 '16 at 12:00
Which downvote? @user236182 – Mad Banners Jul 27 '16 at 23:53

score 2 · Answer 3 · answered Jun 23 '16 at 16:51

$(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta=0\\ (\alpha + \beta) = 6\\ \alpha\beta = 7$

$(x-\alpha - \frac 1\beta)(x-\beta - \frac 1\alpha) = x^2 - (\alpha + \beta + \frac 1\alpha + \frac 1\beta)x + (\alpha\beta + 2+ \frac {1}{\alpha\beta})=0\\ x^2 - (\alpha + \beta + \frac{(\alpha + \beta)}{\alpha\beta})x + (\alpha\beta + 2+ \frac {1}{\alpha\beta})=0\\ x^2 - (6 + \frac 67)x + (9 + \frac 17)=0$

score 2 · Answer 4 · answered Jun 23 '16 at 16:52

$$(x-(\alpha + 1/\beta))(x - (\beta + 1/\alpha)) = x^2 - (\alpha + \beta + 1/\alpha + 1/\beta) x + (\alpha + 1/\beta)(\beta + 1/\alpha $$ You know $\alpha + \beta = 6$ and $\alpha \beta = 7$. $$ \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha \beta} = \dfrac{6}{7}$$ $$\left( \alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) = \alpha \beta + 2 + \frac{1}{\alpha \beta} = 7 + 2 + \frac{1}{7}$$

The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$.

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