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

• May 6th 2013, 04:48 AM
symphonicity
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
• May 6th 2013, 05:05 AM
Prove It
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?
• May 6th 2013, 08:24 AM
Shakarri
Re: Quadratic with 2 variables. Show that it has real roots.
For the equation
$ax^2+bx+c=0$

I'm sure you know that the roots are
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
With this equation you can see that the roots are real if $b^2-4ac$ is not negative.