Question

If $\alpha$ and $\beta$ are the roots of the equation $4x^2-5x+2=0$, find the equation whose roots are

$\alpha +3\beta and3\alpha +\beta$

Answer 1

$$\begin{gathered} \text{Given:}4x^2-5x+2=0 \\ \mathrm{On}\mathrm{comparing}\mathrm{this}\mathrm{equation}\mathrm{with}a{x}^2+bx+c=0,\mathrm{we}\mathrm{get}: \\ \mathit{}a=4,b=-5,c=2 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{\alpha }+\mathrm{\beta }=\frac{-b}{a}\mathrm{and}\mathrm{\alpha \beta }=\frac{c}{a}. \\ \mathrm{Thus},\mathrm{\alpha }+\mathrm{\beta }=-\left({\displaystyle \frac{-5}{4}}\right)=\frac{5}{4}...(1) \end{gathered}$$

$$\begin{gathered} \alpha \beta =\frac{2}{4}=\frac{1}{2}...\left(2\right) \\ \left(i\right)Let{\alpha }_1=\alpha +3\beta and{\beta }_1=3\alpha +\beta \\ Then,{\alpha }_1+{\beta }_1=\alpha +3\beta +3\alpha +\beta \\ =4\alpha +4\beta \\ =4\left(\alpha +\beta \right) \\ =4\times \frac{5}{4}=5(from(1\left)\right) \\ And,{\alpha }_1\times {\beta }_1=\left(\alpha +3\beta \right)\times \left(3\alpha +\beta \right) \\ =3{\alpha }^2+\alpha \beta +9\alpha \beta +3{\beta }^2 \\ =3\left({\alpha }^2+{\beta }^2\right)+10\alpha \beta \\ =3\left[{\left(\alpha +\beta \right)}^2-2\alpha \beta \right]+10\alpha \beta \\ =3{\left(\alpha +\beta \right)}^2+4\alpha \beta \\ =3\times {\left(\frac{5}{4}\right)}^2+4\times \frac{1}{2}(from(2\left)\right) \\ =\frac{75}{16}+2 \\ =\frac{107}{16} \end{gathered}$$


Thus the quadratic equation with roots ${\alpha }_{1}and{\beta }_1$ will be:
$$\begin{gathered} x^2-\left({\alpha }_1+{\beta }_1\right)x+{\alpha }_1{\beta }_1=0 \\ \Rightarrow x^2-\left(5\right)x+\frac{107}{16}=0 \\ \Rightarrow 16x^2-80x+107=0 \end{gathered}$$