Question

Let $a,b\in R-\left\{0\right\}$ and $\alpha ,\beta$ are the roots of $x^2+ax+b=0$, then

• A. $\frac{1}{\alpha },\frac{1}{\beta }$ are roots of $b{x}^2+ax+1=0$
• B. $-\alpha ,-\beta$ are roots of $x^2-ax+b=0$
• C. ${\alpha }^2,{\beta }^2$ are roots of $x^2+(2b-a^2)x+b^2=0$
• D. $\frac{\alpha }{\beta },\frac{\beta }{\alpha }$ are roots of $b{x}^2+(2b-a^2)x+b=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{1}{\alpha },\frac{1}{\beta }$ are roots of $b{x}^2+ax+1=0$
B $-\alpha ,-\beta$ are roots of $x^2-ax+b=0$
C ${\alpha }^2,{\beta }^2$ are roots of $x^2+(2b-a^2)x+b^2=0$
D $\frac{\alpha }{\beta },\frac{\beta }{\alpha }$ are roots of $b{x}^2+(2b-a^2)x+b=0$

$\alpha ,\beta$ are roots of $x^2+ax+b=0$
$\Rightarrow \alpha +\beta =-a$ and $\alpha \beta =b$

$A)$
For $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ as roots
$\frac{1}{\alpha }+\frac{1}{\beta }=-\frac{a}{b}$ and $\frac{1}{\alpha }\frac{1}{\beta }=\frac{1}{b}$
$\Rightarrow$ Equation is $x^2-(-\frac{a}{b})x+\frac{1}{b}=0$
$\Rightarrow b{x}^2+ax+1=0$

$B)$
For $-\alpha$ and $-\beta$ as roots
$-\alpha -\beta =a$ and $(-\alpha )(-\beta )=b$
$\Rightarrow$ Equation is $x^2-ax+b=0$

$C)$
For ${\alpha }^2$ and ${\beta }^2$ as roots
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =a^2-2b$
and ${\alpha }^2{\beta }^2=b^2$
$\Rightarrow$ Equation is $x^2-(a^2-2b)x+b^2=0$
$\Rightarrow x^2+(2b-a^2)x+b^2=0$

$D)$
For $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as roots,
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{a^2-2b}{b}$
and $\frac{\alpha }{\beta }\frac{\beta }{\alpha }=1$
$\Rightarrow$ Equation is $x^2-\left(\frac{a^2-2b}{b}\right)x+1=0$
$\Rightarrow b{x}^2+(2b-a^2)x+b=0$