Question

If $\alpha$ and $\beta$ are the roots of the equation $a{x}^2+bx+c=0$ and $p{x}^2+qx+r=0$ has roots$\frac{1-\alpha }{\alpha }$ and $\frac{1-\beta }{\beta }$, then $r=?$

• A. $a+2b$
• B. $a+b+c$
• C. $ab+bc+ca$
• D. $abc$

Answer 1

The correct option is B

$a+b+c$

Explanation for the correct option:

Step 1: Find the sum and product of the roots of first quadratic equation.

We have given two quadratic equations$a{x}^2+bx+c=0$ and $p{x}^2+qx+r=0$, the roots of first equation are $\alpha$ and $\beta$ the roots of second equation are$\frac{1-\alpha }{\alpha }$ and$\frac{1-\beta }{\beta }$.

and we want to find $r$

For equation $a{x}^2+bx+c=0$,

$\alpha +\beta =\frac{-b}{a}$

$\alpha \beta =\frac{c}{a}$

Step 2: Find the sum of the roots of second quadratic equation.

For equation $p{x}^2+qx+r=0$

$\begin{array}{rcl}\frac{1-\alpha }{\alpha }+\frac{1-\beta }{\beta }& =& \frac{-q}{p}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{\beta (1-\alpha )+\alpha (1-\beta )}{\alpha \beta }& =& \frac{-q}{p}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{\beta -\alpha \beta +\alpha -\alpha \beta }{\alpha \beta }& =& \frac{-q}{p}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{{\displaystyle \raisebox{1ex}{$-b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.-2\left(\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)}}{\left({\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}\right)}& =& \frac{-q}{p}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{-(b+2c)}{c}& =& -\frac{q}{p}.....\left(1\right)\end{array}$

Step 3: Find the product of the roots of second quadratic equation.

$\begin{array}{rcl}\left(\frac{1-\alpha }{\alpha }\right)\left(\frac{1-\beta }{\beta }\right)& =& \frac{r}{p}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\left(\frac{1-\beta -\alpha +\alpha \beta }{\alpha \beta }\right)& =& \frac{r}{p}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1-(\beta +\alpha )+\alpha \beta }{\alpha \beta }& =& \frac{r}{p}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1+({\displaystyle \raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.)}+({\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$a$}\right.)}}{\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}& =& \frac{r}{p}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{a+b+c}{c}& =& \frac{r}{p}.......\left(2\right)\end{array}$

Step 4: By comparing the equations (1) and (2), we get

$p=c,q=b+2c$ and

$\mathit{r}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{c}$

Hence, the correct option is$\left(B\right)$