Question

Let$\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0$ and ${\alpha }_1,-\beta$be the roots of $a_1{x}^2+b_{1}x+c_1=0$ then $\alpha ,{\alpha }_1$ are the roots of

• A. $\frac{x^2}{\left[\left(\frac{b}{a}\right)+\left(\frac{b_1}{a_1}\right)\right]}+x+\frac{1}{\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right]}=0$
• B. $\frac{x^2}{\left[\left(\frac{b}{a}\right)+\left(\frac{b_1}{a_1}\right)\right]}+\frac{x}{\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right]}+1=0$
• C. $\frac{x^2}{\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right]}+x+\frac{1}{\left[\left(\frac{b}{a}\right)+\left(\frac{b_1}{a_1}\right)\right]}=0$
• D. $\frac{x^2}{\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right]}+\frac{1}{\left[\left(\frac{b}{a}\right)+\left(\frac{b_1}{a_1}\right)\right]}+1=0$

Answer 1

The correct option is A

$\frac{x^2}{\left[\left(\frac{b}{a}\right)+\left(\frac{b_1}{a_1}\right)\right]}+x+\frac{1}{\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right]}=0$

Explanation for the correct option:

Finding the equation whose roots are $\alpha ,{\alpha }_1$:

$\alpha$ and$\beta$ be the roots of $a{x}^2+bx+c=0$

Sum of roots $\alpha +\beta =-\frac{b}{a}....\left(i\right)$

Product of roots $\alpha \beta =\frac{c}{a}...\left(ii\right)$

${\alpha }_1,-\beta$ be the roots of $a_1{x}^2+b_{1}x+c_1=0$

$$\begin{gathered} {ɑ}_1+(-\beta )=\frac{-b_1}{a_1}.....\left(iii\right) \\ {ɑ}_1(-\beta )=\frac{c_1}{a_1}.....\left(iv\right) \end{gathered}$$

Adding equations $\left(i\right)\&\left(iii\right)$

$$\begin{gathered} \Rightarrow (ɑ+\beta )+({ɑ}_1–{\beta }_1)=\frac{-b}{a}–\frac{b_1}{a_1} \\ \Rightarrow ɑ+{ɑ}_1=–\left[\frac{b}{a}+\frac{b_1}{a_1}\right]........\left(v\right) \end{gathered}$$

Dividing equation $\left(i\right)$ by equation $\left(ii\right)$

$$\begin{gathered} \Rightarrow \frac{(\alpha +\beta )}{\alpha \beta }=\frac{{\displaystyle \frac{-b}{a}}}{{\displaystyle \frac{c}{a}}} \\ \Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }=–\frac{b}{c}.....\left(vi\right) \end{gathered}$$

Similarly dividing equation $\left(iii\right)$ by equation $\left(iv\right)$

$\Rightarrow \frac{1}{\alpha }+\left(\frac{-1}{\beta }\right)=–\frac{b_1}{c_1}.....\left(vii\right)$

Now adding equations $\left(vi\right)\&\left(vii\right)$

$$\begin{gathered} \Rightarrow \frac{1}{\alpha }+\frac{1}{{\alpha }_1}=-\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right] \\ \Rightarrow \frac{\alpha +{\alpha }_1}{\alpha {\alpha }_1}=–\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right]....\left(viii\right) \end{gathered}$$
The equation whose roots are $\alpha ,-{\alpha }_1$ is $x^2–(\alpha +{\alpha }_1)x+\alpha {\alpha }_1=0....\left(ix\right)$

Dividing $(\alpha +{\alpha }_1)$ in equation $\left(ix\right)$ on both sides

$$\begin{gathered} \Rightarrow \frac{x^2}{(\alpha +{\alpha }_1)}–x+\frac{\alpha {\alpha }_1}{(\alpha +{\alpha }_1)}=0 \\ \Rightarrow \frac{x^2}{\left[\left(\frac{b}{a}\right)+\left(\frac{b_1}{a_1}\right)\right]}+x+\frac{1}{\left[\left(\frac{b}{c}\right)+\left(\frac{b_1}{c_1}\right)\right]}=0\text{[from equations}\left(v\right)\&\left(viii\right)] \end{gathered}$$

Hence, option (A) is correct.