Roots

Algebra Level 2

If $\alpha,\beta$ are the roots of the equation $x^2+ax+b=0$ ,where $b \neq 0$ ,then the roots of the equation $bx^2+ax+1=0$ are

\alpha^2,\beta^2

1/\alpha,1/\beta

1/\alpha^2,1/\beta^2

\alpha/\beta,\beta/\alpha

1 solution

Tijmen Veltman
Jun 9, 2015

We have $x^2+ax+b=(x-\alpha)(x-\beta)$ , meaning that $-a=\alpha+\beta$ and $b=\alpha\beta$ (Vieta's law). This gives us:

$bx^2+ax+1=b(x^2+\frac{a}{b}x+\frac1{b}) =b(x^2-\frac{\alpha+\beta}{\alpha\beta}x+\frac1{\alpha\beta})$

$=b(x^2-(\frac1\alpha+\frac1\beta)x+\frac1\alpha\frac1\beta) =b(x-\frac1\alpha)(x-\frac1\beta)=0$ .

Therefore, the roots are $\boxed{1/\alpha,1/\beta}$ .