If alpha and beta are multiple zeroes of 6x^2-5x-k and alpha-beta=7/6

1.	If alpha and beta are multiple zeroes of 6x^2-5x-k and alpha-beta=7/6 so find the value of k
Answer» $\large{\underline{\underline{\mathfrak{\green{\sf{QUESTION:-}}}}}}$ . if $\alpha\:and\:\beta$ are MULTIPLE ZEROES of $\:(6x^<klux>2</klux>-5x-k)$ and $\alpha\:-\:\beta\:=\:\frac{7}{6}$ so find the value of k $\large{\underline{\underline{\mathfrak{\green{\sf{Sulution:-}}}}}}$ . $\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$ . Equation is $\:(6x^2-5x-k)\:=\:0$ $\alpha\:and\:\beta\:are\:zeros\:of\:this\:equation$ $\alpha\:-\:\beta\:=\frac{7}{6}$ ...(1) $\large{\underline{\underline{\mathfrak{\sf{Find\:Here:-}}}}}$ . Value of k $\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$ . We KNOW , $\bold{\boxed{\boxed{\:Sum\:of\:Zeros\:=\frac{-(Coefficient\:of\:x)}{(Coefficient\:of\:x^2)}}}}$ So, $\implies\:(\alpha\:+\beta)\:=\frac{-(-5)}{6}$ $\implies\:(\alpha\:+\beta)\:=\frac{5}{6}$ ...(2) ADDITION of equation (1) and (2) , We find here, $\implies\:2\alpha\:=\frac{7}{6}\:+\frac{5}{6}$ $\implies\:2\alpha\:=\frac{(7+5)}{6}$ $\implies\:2\alpha\:=\frac{\cancel{12}}{\cancel{6}}$ $\bold{\boxed{\boxed{\:\alpha\:=\:1}}}$ Keep value of $\alpha\:=\:1$ in (2) we find here, $\implies\:(1\:+\beta)\:=\frac{5}{6}$ $\implies\:(\beta)\:=\frac{5}{6}\:-\:1$ $\implies\:(\beta)\:=\frac{(5-6)}{6}$ $\bold{\boxed{\boxed{\:(\beta)\:=\frac{-1}{6}}}}$ ___________________ Again, We know , $\bold{\boxed{\boxed{\:Product\:of\:Zeros\:=\frac{(Constant\:part)}{(Coefficient\:of\:x^2)}}}}$ $\implies\:(\alpha\times\beta)\:=\frac{(-k)}{6}$ Keep value of $\alpha\:and\:\beta$ $\implies\:(1\:\times\frac{-1}{6})\:=\frac{k}{6}$ $\implies\:\frac{k}{6}\:=\frac{-1}{6}$ $\implies\:(k)\:=\frac{\cancel{-6}}{\cancel{6}}$ $\bold{\boxed{\boxed{\:(k)\:=\:-1}}}$ _____________________ $\large{\underline{\underline{\mathfrak{\green{\sf{Answer:-}}}}}}$ . Value of k = -1 ___________________