Find the Values of k for Which the Equation Has Equal Roots
Solution
Given: $$2x^2+kx+3=0$$
Here, $$a=2,\quad b=k,\quad c=3$$
For equal roots, $$D=b^2-4ac=0$$
$$k^2-4(2)(3)=0$$
$$k^2-24=0$$
$$k=\pm\sqrt{24}=\pm2\sqrt6$$
Answer
The value(s) of k for which the roots are equal is: $$\boxed{k=\pm2\sqrt6}$$