Question

Choose the quadratic equation(s) whose difference and product of roots are given as $3$ and $28,$ respectively.

• A. $x^2-x+28=0$
• B. $x^2-11x+28=0$
• C. $3x^2+x+28=0$
• D. $x^2+11x+28=0$

Answer 1

Answer: (B), (D)

$x^2+11x+28=0$

Let $\alpha$ and $\beta$ be the two roots of the required quadratic equation $a{x}^2+bx+c=0$ , where $\alpha >\beta$.
We know $\alpha \beta =\frac{c}{a}$
Also, $(\alpha -\beta )=\frac{\sqrt{D}}{|a|}$
Where $D=b^2-4ac$
Now checking with each options:
$1$. $x^2-28x+11=0$
Here $a=1,b=-28,c=11$
So difference of roots will be
$\alpha -\beta =\frac{\sqrt{740}}{1}$
$\alpha -\beta \ne 3$
So this option is not correct .
Similarly for option $2$
$x^2-11x+28=0$
Here $a=1,b=-11,c=28$
So difference of roots will be
$\alpha -\beta =\frac{\sqrt{9}}{1}$
$\alpha -\beta =3$
Also, $\alpha \beta =\frac{c}{a}$
$\alpha \beta =28$
So this option is correct.


Similarly for option $3.$
$x^2+11x+28=0$
Here $a=1,b=11,c=28$
So difference of roots will be
$\alpha -\beta =\frac{\sqrt{9}}{1}$
$\alpha -\beta =3$
Also, $\alpha \beta =\frac{c}{a}$
$\alpha \beta =28$
So this option is correct.


Similarly for option $4$
$3x^2+11x+10=0$
Here $a=3,b=11,c=10$
So difference of roots will be
$\alpha -\beta =\frac{\sqrt{1}}{3}$
$\alpha -\beta =\frac{1}{3}$
$\alpha -\beta \ne 3$
So this option is not correct.