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A student while solving a quadratic equation in $x$, he copied its constant term incorrectly and got its roots as 5 and 9. Another student copied the constant term and coefficient of $x^2$ of the same equation correctly as 12 and 4 respectively. If $s, p$ and $\Delta$ denote respectively the sum of the roots, the product of the roots and the discriminate of the correct equation, then $\frac{\Delta}{3 p+s}=$
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Verified Answer
The correct answer is:
128
Let the equation be $a x^2+b x+c=0$
Sum of roots $=-\frac{b}{a}$
and product of roots $=\frac{c}{a}$
when constant term is copied wrongly we get roots as 5 and 9.
$\therefore \quad-\frac{b}{a}=5+9 \quad$ [sum will remain correct]
$\Rightarrow \quad-\frac{b}{a}=14$
Another student copied constant term and coefficient of $x^2$ correctly us 12 and 4 .
$$
\begin{array}{ll}
\therefore & c=12 \text { and } a=4 \\
\therefore & \frac{-b}{a}=14 \Rightarrow b=-56 \\
\therefore & S=-\frac{b}{a}=\frac{-(-56)}{4}=14
\end{array}
$$
$$
\begin{aligned}
P & =\frac{c}{a}=\frac{12}{4}=3 \\
\Delta & =b^2-4 a c \\
& =(-56)^2-4 \times 4 \times 12 \\
& =3136-192=2944 \\
\therefore \frac{\Delta}{3 p+S} & =\frac{2944}{3 \times 3+14}=\frac{2944}{23}=128
\end{aligned}
$$
Sum of roots $=-\frac{b}{a}$
and product of roots $=\frac{c}{a}$
when constant term is copied wrongly we get roots as 5 and 9.
$\therefore \quad-\frac{b}{a}=5+9 \quad$ [sum will remain correct]
$\Rightarrow \quad-\frac{b}{a}=14$
Another student copied constant term and coefficient of $x^2$ correctly us 12 and 4 .
$$
\begin{array}{ll}
\therefore & c=12 \text { and } a=4 \\
\therefore & \frac{-b}{a}=14 \Rightarrow b=-56 \\
\therefore & S=-\frac{b}{a}=\frac{-(-56)}{4}=14
\end{array}
$$
$$
\begin{aligned}
P & =\frac{c}{a}=\frac{12}{4}=3 \\
\Delta & =b^2-4 a c \\
& =(-56)^2-4 \times 4 \times 12 \\
& =3136-192=2944 \\
\therefore \frac{\Delta}{3 p+S} & =\frac{2944}{3 \times 3+14}=\frac{2944}{23}=128
\end{aligned}
$$
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