This is only one variable (x), while the other letter is a parameter, so it and any combinations of it are treated as constants.
Now to have real roots, the discriminant is nonnegative. Can you write an equation for that situation?
Hi everyone,
I'm really stumped on this question. Would really appreciate some guidance... "show that 3vx^2 + (4v-6)x + 2v-5 = 0 has real solutions in x if and only if (v-3)(2v-3)<=0" (lesser than or equal to zero).
Thank you to anyone who can help...
symphonicity
This is only one variable (x), while the other letter is a parameter, so it and any combinations of it are treated as constants.
Now to have real roots, the discriminant is nonnegative. Can you write an equation for that situation?