Maths Question:
$\begin{align}
& \text{The coefficient of }x\text{ in quadratic equation }{{x}^{2}}+px+q\text{ was taken as 17 in place of 13 } \\
& \text{and its roots were found to }-2\text{ and }-15.\text{ Find the roots of the original equation} \\
\end{align}$
Maths Solution:
$\begin{align}
& \text{Given }{{x}^{2}}+px+q=0 \\
& \text{The coefficient of }x\,\text{is 7 instead of 13} \\
& \text{the roots of which are }-2\text{ and }-5 \\
& \text{i}\text{.e}\text{. (}x+2)(x+15)=0 \\
^{2}}+17x+30=0\text{ meaning }q=30 \\
& \text{The original equation is} \\
^{2}}+13x+30=0 \\
& (x+3)(x+10)=0 \\
& \text{The roots of the original equation are } \\
& x=-3\text{ and }x=-10 \\
\end{align}$
University mathstopic: