Quadratic with 2 variables. Show that it has real roots.

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

Re: Quadratic with 2 variables. Show that it has real roots.

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?

Re: Quadratic with 2 variables. Show that it has real roots.

For the equation

$\displaystyle ax^2+bx+c=0$

I'm sure you know that the roots are

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

With this equation you can see that the roots are real if $\displaystyle b^2-4ac$ is not negative.