Question

Question 4

Answer 1

Answer:

No real roots

Step-by-step explanation:

The given equations are:

$3x-1+2y=0$

$3x^2-y^2+4=0$

We make y the subject in the first equation to get

$y=\frac{1-3x}{2}$

We substitute into the second expression to get:

$3x^2-(\frac{1-3x}{2})^2+4=0$

We expand to get:

$3x^2-\frac{(1-6x+9x^2)}{4}+4=0$

Multiply through by 4 to get:

$12x^2-1+6x-9x^2+16=0$

$3x^2+6x+15=0$

The discriminant is given by $D=b^2-4ac$

$D=3^2-4*12*15$

$D=-711$

Since the discriminant is less than zero, the two curves never intersects.

Therefore the system has no real roots