If roots of the quadratic equation
3y²+k y +12=0
are real and equal
then find k​

Let [tex]a=3[/tex], [tex]b=k[/tex], [tex]c=12[/tex] in [tex]3y^2+ky+12=0[/tex] .

we are given that [tex]b^2-4ac=0[/tex].

so, [tex]k^2-2(3)(12)=0[/tex]

now, [tex]k^2=6*6*2=6^2*2[/tex]

[tex]k=6\sqrt{2}[/tex]

∴ equation is as follows:-

[tex]3y^2+6\sqrt{2} y+12=0[/tex]