Question

Question 4

Find the values of x and y that satisfy 3x - 1 + 2y = 0 and 3x2 - y2 + 4 = 0. Use the algebraic method to find the solution, and then use the graphing method to confirm whether the system of equations has real roots.

Answer 1

No real roots

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=(1-3x)/(2)`

We substitute into the second expression to get:

`3x^2-((1-3x)/(2))^2+4=0`

We expand to get:

`3x^2-((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