Answer:
No real roots
Step-by-step explanation:
The given equations are:
[tex]3x-1+2y=0[/tex]
[tex]3x^2-y^2+4=0[/tex]
We make y the subject in the first equation to get
[tex]y=\frac{1-3x}{2}[/tex]
We substitute into the second expression to get:
[tex]3x^2-(\frac{1-3x}{2})^2+4=0[/tex]
We expand to get:
[tex]3x^2-\frac{(1-6x+9x^2)}{4}+4=0[/tex]
Multiply through by 4 to get:
[tex]12x^2-1+6x-9x^2+16=0[/tex]
[tex]3x^2+6x+15=0[/tex]
The discriminant is given by [tex]D=b^2-4ac[/tex]
[tex]D=3^2-4*12*15[/tex]
[tex]D=-711[/tex]
Since the discriminant is less than zero, the two curves never intersects.
Therefore the system has no real roots
