Question

If $p,q$ are the roots of the equation $3x^2+7x-2=0,$ then the equation whose roots are $\frac{p}{5}+3,\frac{q}{5}+3$ is $a{x}^2+bx+c=0.$ Now the value of $\frac{c-b}{a}=\frac{k}{75},$ then $k$ is

• A. 983.0
• B. 983.00
• C. 983
• D. 0983

Answer 1

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

Given: $3x^2+7x-2=0;$ roots are $p,q$

$\text{Sum of roots}=p+q=\frac{-7}{3}\cdots \left(i\right)$

$\text{Product of roots}=p.q=\frac{-2}{3}\cdots \left(ii\right)$

Now we have to find the equation whose roots are $\frac{p}{5}+3,\frac{q}{5}+3$
$\therefore \text{Sum of roots}=\frac{p}{5}+3+\frac{q}{5}+3=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=6+(\frac{p+q}{5})$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=6+(\frac{(\frac{-7}{3})}{5})\left[\text{From}\right(i\left)\right]$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=(\frac{83}{15})$

Now, $\text{Product of roots}=(\frac{p}{5}+3)(\frac{q}{5}+3)=\frac{c}{a}$
$\Rightarrow \text{Product of roots}=\frac{c}{a}=\frac{p.q}{25}+\frac{3(p+q)}{5}+9$

$\Rightarrow \text{Product of roots}=\frac{c}{a}=\frac{(\frac{-2}{3})}{25}+\frac{3(\frac{-7}{3})}{5}+9\left[\text{From}\right(i),(ii\left)\right]$

$\Rightarrow \text{Product of roots}=\frac{c}{a}=(\frac{568}{75})$

$\therefore \frac{c-b}{a}=(\frac{c}{a})+(\frac{-b}{a})$

$\Rightarrow \frac{c-b}{a}=(\frac{568}{75})+(\frac{83}{15})$

$\Rightarrow \frac{c-b}{a}=\frac{983}{75}$