Complex Numbers...

Find the value of $$k$$ for which the equation $$(k+1){x}^{2}-2(k-1)x+1=0$$ has real and equal roots

The quadratic equations $$\displaystyle x^{2}-5x+3=0$$ has 

If the roots of the equation $$\displaystyle x^{2}+px-6=0$$ are $$6$$ and $$-1$$ then the value of $$p$$ is

The roots of the equation, where $$\displaystyle a\: \: \epsilon \: \: R$$  are 

If the roots of the equation $$\displaystyle (a^{2}+b^{2})x^{2}-2b(a+c)x+(b^{2}+c^{2})=0$$ are equal then =

If the equation $$\displaystyle x^{2}-2kx-2x+k^{2}=0 $$ has equal roots the value of k must be

For what value of k will $$\displaystyle x^{2}-\left ( 3k-1 \right )x+2k^{2}+2k=11$$ have equal roots?

Find the value of $$k$$ for which the equation $${x}^{2}-6x+k=0$$ has distinct roots.

If $${z}_{1},{z}_{2}$$ are two complex numbers and $$c>0$$ such that $${ \left| { z }_{ 1 }+{ z }_{ 2 } \right|  }^{ 2 }\le \left( 1+c \right) { \left| { z }_{ 1 } \right|  }^{ 2 }+k{ \left| { z }_{ 2 } \right|  }^{ 2 },$$ then $$k=$$