Quadratic Equations

If $\alpha \neq 0$ and 0 are the roots of$x^2-5k x+(6k^2-2k)=0$, then find $\alpha$.Step 1: Use the fact that $0$ is a root.If $x=0$ is a root, it should satisfy the equation:$\;0^2-5k(0)+(6k^2-2k)=0$$\; \Rightarrow 6k^2-2k=0$$\; \Rightarrow 2k(3k-1)=0$So $k=0$ or $k=\frac{1}{3}$.But if $k=0$, the equation becomes $x^2=0$, giving a double root 0 - contradicting that one root is nonzero $(\alpha \neq 0)$.Hence,$k=\frac{1}{3}.$Step 2: Substitute $k=\frac{1}{3}$ into the equation.$\;x^2-5\left(\frac{1}{3}\right)x+\left(6\left(\frac{1}{3}\right)^2-2\left(\frac{1}{3}\right)\right)=0$$\; \Rightarrow x^2-\frac{5}{3}x+\left(\frac{6}{9}-\frac{2}{3}\right)=0$$\; \Rightarrow x^2-\frac{5}{3}x+\left(\frac{2}{3}-\frac{2}{3}\right)=0$Wait - compute the constant term carefully:$\; \frac{6}{9}-\frac{2}{3}=\frac{2}{3}-\frac{2}{3}=0.$So the equation becomes:$\; x^2-\frac{5}{3}x=0$$\; x\left(x-\frac{5}{3}\right)=0$Step 3: Identify the roots.Roots are $x=0$ and $x=\frac{5}{3}$.Given that one root is 0 and another is $\alpha \neq 0$,$\alpha=\frac{5}{3}.$Final Answer:$\alpha=\frac{5}{3}$