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If $\alpha \neq \beta$ but $\alpha^{2}=5 \alpha-3$ and $\beta^{2}=5 \beta-3$ then the equation having $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ as its roots is

1) $3 x^{2}-19 x+3=0$
2) $3 x^{2}+19 x-3=0$
3) $3 x^{2}-19 x-3=0$
4) $x^{2}-5 x+3=0$
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(1)

Explanation

We have $\alpha^{2}=5 \alpha-3$ and $\beta^{2}=5 \beta-3$; $\Rightarrow \alpha \& \beta$ are roots of equation, $x^{2}=5 x-3$ or

$x^{2}-5 x+3=0$

$\therefore \alpha+\beta=5$ and $\alpha \beta=3$

Thus, the equation having $\frac{\alpha}{\beta} \& \frac{\beta}{\alpha}$ as its roots is $x^{2}-x\left(\frac{\alpha}{\beta}+\frac{\beta} {\alpha}\right)+\frac{\alpha \beta}{\alpha \beta}=0$

$\Rightarrow x^{2}-x\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right)+1=0$ OR $3 x^{2}-19 x+3=0$

by Diamond (75,914 points)

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