Question

If $\alpha ,\beta$ are the corresponding roots of the given quadratic equations. Then match the following.

• A. $\alpha ,\beta <0$
• B. $\alpha \beta <0$
• C. $\alpha =\frac{1}{\beta }$
• D. $\alpha =0,\beta \ne 0$

Answer 1

Using the concept of nature of roots of a quadratic equation on the basis of coefficients, we can easily get the solution.
Let $\alpha ,\beta$ be the roots of the given equations.
$\text{for}a{x}^2+bx+c=0;a,b,c\in R\text{and}a\ne 0$
$\text{we have the following results}$
$\left(i\right)a=c\Rightarrow \text{roots are reciprocal to each other}\left(ii\right)b=0\text{and sign of a,c are different sign}\Rightarrow \text{roots are}\pm \sqrt{\frac{-c}{a}}\left(iii\right)a,b,c\text{are of same sign}\Rightarrow \text{Both roots are negative}\left(iv\right)c=0\Rightarrow \text{roots are 0,}\frac{-b}{a}$

Hence we can conclude the following:
$A.3x^2+10x+3=0$
Here, $a=c\ne b,$ thus the roots will be reciprocal to each other
Or $\alpha =\frac{1}{\beta }$
$B.5x^2-125=0$
Here $b=0\Rightarrow$ The roots will have opposite signs.
Or $\alpha \beta <0$
$C.x^2+5x+6=0$
Here, All the coefficients have same sign.
$\therefore$ The roots will be negative.
Or $\alpha ,\beta <0$
$D.3x^2+9x=0$
Here, $c=0\Rightarrow$ The roots will be $0,-\frac{b}{a}$
$\Rightarrow$ The roots will be $0,-3$
Or $\alpha =0,\beta \ne 0$