Question

# If the roots of the quadratic equation $$kx^2+(a+b)x+ab=0$$ are $$(-1, -b)$$, find the value of k.

Solution

## $$\begin{matrix} Roots\, are\left( { -1,\, -b } \right) ,\, then\, k=? \\ Then,\, satisfy\, the\, \, roots\, in\, equation \\ \Rightarrow k{ \left( { -1 } \right) ^{ 2 } }+\left( { a+b } \right) \left( { -1 } \right) +ab=0 \\ \Rightarrow k-\left( { a+b } \right) +ab=0 \\ \Rightarrow k=\left( { a+b } \right) -ab....\left( 1 \right) \\ again\, put\, x=-b \\ \Rightarrow k{ \left( { -b } \right) ^{ 2 } }+\left( { a+b } \right) \left( { -b } \right) +ab=0 \\ \Rightarrow k{ b^{ 2 } }-{ b^{ 2 } }-ab+ab=0 \\ \Rightarrow k{ b^{ 2 } }-{ b^{ 2 } }=0 \\ \Rightarrow k{ b^{ 2 } }={ b^{ 2 } } \\ k=1 \\ \end{matrix}$$Maths

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