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The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.

The answer is $4x^2+5x+4=0$

I don't know how to get to the answer. Could someone explain the steps?

nonuser
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Anne
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6 Answers6

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Rename $a= \alpha $ and $b=\beta$

So this is $(x-{a\over b})(x-{b\over a})=0$ thus

$$x^2-{a^2+b^2\over ab}x+1=0$$

Since $ab = {6\over 2}=3$ and $a^2+b^2 = (a+b)^2-2ab = ({3\over 2})^2-6 = {-15\over 4} $ we get:

$$x^2+{5\over 4}x+1=0$$

and thus the conclusion.

nonuser
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By Vieta's theorem we know that $\alpha+\beta=\frac{3}{2}$ and $\alpha\beta=3$. If follows that

$$ \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{(\alpha+\beta)^2}{\alpha\beta}-2 =\frac{3}{4}-2=-\frac{5}{4}$$ and obviously $\frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha}=1$. It follows that a polynomial vanishing at $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is given by $z^2+\frac{5}{4}z+1$ or by $\color{red}{4x^2+5x+4}$ as stated.

Jack D'Aurizio
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Hint:

First approach: Find $\alpha$ and $\beta$ first.

Second approach: $$\frac{\alpha}{\beta} + \frac{\beta}{\alpha}=\frac{(\alpha^2+\beta^2)}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$$

Siong Thye Goh
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$$\alpha'=\frac{\alpha}{\beta} ,\beta'=\frac{\beta}{\alpha}$$ now find sum and product of $\bf new$ roots $$\quad{S'=\alpha'+\beta'\\p'=\alpha'\times \beta'\\ S'=\alpha'+\beta'=\frac{\alpha}{\beta} +\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{s^2-2p}{p}=\frac{(-\frac{-3}{2})^2-2.\frac{6}{2}}{\frac{6}{2}}\\p'=\alpha'\times\beta'=\frac{\alpha}{\beta} \times\frac{\beta}{\alpha}=1}$$ then put into $$\quad{x^2-s'x+p'=0\\x^2-\frac{-5}{4}x+1=0 \to \times 4 \\4x^2+5x+4=0}$$

Khosrotash
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As $\alpha+\beta=\dfrac32, \alpha\beta=\dfrac62$

let $y=\dfrac\alpha\beta\iff y+1=\dfrac3{2\beta}\iff\beta=\dfrac3{2(y+1)}$

But as $\beta$ is a root of $$2x^2-3x+6=0$$

$$2\left(\dfrac3{2(y+1)}\right)^2-3\left(\dfrac3{2(y+1)}\right)+6=0$$

As $y+1\ne0,$ multiply both sides by $\dfrac{2(y+1)^2}3$ to find $$0=3-3(y+1)+4(y+1)^2=4y^2+5y+4$$

By symmetry, we can surmise that the same equation will be reached if we start with $y=\dfrac\beta\alpha$

lab bhattacharjee
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  • See also : https://math.stackexchange.com/questions/1837156/the-roots-of-the-equation-x2-6x-7-0-are-%CE%B1-and-%CE%B2-find-the-equation/1837170 https://math.stackexchange.com/questions/1909362/product-of-one-minus-the-tenth-roots-of-unity/1909366#1909366 https://math.stackexchange.com/questions/1909362/product-of-one-minus-the-tenth-roots-of-unity https://math.stackexchange.com/questions/1811081/problem-based-on-sum-of-reciprocal-of-nth-roots-of-unity – lab bhattacharjee Sep 26 '17 at 18:02
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$$ \left( x - \frac \alpha \beta \right)\left( x - \frac\beta\alpha\right) = 0 $$ $$ (\beta x-\alpha)(\alpha x-\beta) = 0 $$ $$ \alpha\beta x^2 - (\alpha^2+\beta^2) x + \alpha \beta = 0 \tag 1 $$

If $2x^2-3x+6=0$ then $x = \dfrac{3 \pm \sqrt{-39}} 4.$ So $$ \alpha\beta = \frac{3+\sqrt{-39}} 4 \cdot \frac{3-\sqrt{-39}} 4 = 3. $$ $$ \alpha^2+\beta^2 = \left( \frac{3+\sqrt{-39}} 4 \right)^2 + \left( \frac{3-\sqrt{-39}} 4 \right)^2 = \frac{-30}{16} + \frac{-30}{16} = \frac{-15} 4. $$ So line $(1)$ becomes $$ 3x^2 + \frac{15} 4 x + 3 = 0 $$ which is equivalent to $$ 4x^2 + 5x + 4 =0. $$

Michael Hardy
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