Finding the equation whose roots are

Question

Okay I found this one on a test and it must be something really silly I am missing $ \alpha$ is not equal to $\beta$ and $\alpha^2=5\alpha-3,5\beta-3=\beta^2$ then the equation whose roots are $\alpha/\beta$ and $ \beta/\alpha$ is $ (a)3x^2-25+3=0,(b)x^2-5x+3=0,(c)x^2+5x-3=0,(d)3x^2-19x+3=0$ Of course the answer is either a or d I tried solving for $\alpha$ and $\beta$ but the equation became too complicated

Answer 1

HINT:

Method$\#1:$

So, $\alpha, \beta$ are the unequal roots of $$t^2-5t+3=0$$

$$\alpha+\beta=?, \alpha\cdot\beta=?$$

Method$\#2:$

If $a^2=5a-3, b^2=5b-3$

$a^2-b^2=5(a-b)\iff a+b=5$ as $a-b\ne0$

$a^2+b^2=5(a+b)-6,5^2-2ab=5\cdot5-6\iff ab=3$

Now from any one of the above methods,

$$\dfrac\alpha\beta+\dfrac\beta\alpha=\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{(\alpha+\beta)^2}{\alpha\beta}-2$$

$$\dfrac\alpha\beta\cdot\dfrac\beta\alpha=1$$

Answer 2

Notice, the roots $\alpha$ & $\beta$ are distinct roots & $$\alpha^2=5\alpha -3, \ \ 5\beta-3=\beta^2$$

Hence, the quadratic equation is $$x^2-5x+3=0$$ whose roots are determined by using quadratic formula $$\alpha, \beta=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-3)}}{2(1)}$$ $$=\frac{5\pm\sqrt{37}}{2}$$ $$\implies \alpha=\frac{5+\sqrt{37}}{2}, \ \ \beta=\frac{5-\sqrt{37}}{2}$$ Now, $$\frac{\alpha}{\beta}=\frac{\frac{5+\sqrt{37}}{2}}{\frac{5-\sqrt{37}}{2}}=-\frac{31+5\sqrt {37}}{6}$$ $$\frac{\beta}{\alpha}=\frac{\frac{5-\sqrt{37}}{2}}{\frac{5+\sqrt{37}}{2}}=-\frac{31+5\sqrt {37}}{6}=-\frac{31-5\sqrt {37}}{6}$$

Hence, the quadratic equation having roots $\frac{\alpha}{\beta}$ & $\frac{\beta}{\alpha}$ is given as $$\left(x-\frac{\alpha}{\beta}\right)\left(x-\frac{\beta}{\alpha}\right)=0$$
$$\left(x+\frac{31+5\sqrt {37}}{6}\right)\left(x+\frac{31-5\sqrt {37}}{6}\right)=0$$ $$\color{red}{3x^2+31x+3=0}$$