# f(x) = 3x^2-3x+1 g(x) = 2x^2-4x-3?

Determine whether the following systems of quadratic equations intersect and how many points they intersect at.

### 6 Answers

- PhilipLv 67 months ago
f(x) = 3x^2-3x+1, g(x) = 2x^2-4x-3.

Put h(x) = f(x)-g(x). Then h(x) = x^2+x+4.

Discriminant of h(x) = 1^2 -4*1*4 = -15, < 0. Therefore h(x) has no real zeros. f(x) & g(x) do not intersect.

- JohnathanLv 77 months ago
Set the functions equal.

3x^2 - 3x + 1 = 2x^2 - 4x - 3

x^2 + x + 4 = 0

Use the discriminant of the quadratic formula, and...

1^2 - 4(1)(4) = 1 - 16 = -15

Negative discriminant means imaginary roots/solutions. These quadratics won't ever intersect.

- PopeLv 77 months ago
Those are two quadratic function definitions, not a system of quadratic equations. I think you must be asking about the number of intersections of the curves that correspond to these two equations:

y = f(x)

y = g(x)

The intersections (if any) are solutions to this system of second-degree equations:

y = 3x² - 3x + 1

y = 2x² - 4x - 3

3x² - 3x + 1 = 2x² - 4x - 3

x² + x + 4 = 0

That, now, is a quadratic equation. The x-coordinate of any point of intersection would have to satisfy this equation.

discriminant = (1)² - 4(1)(4) = -15

discriminant < 0

For a discriminant less than zero there can be no real solution to the equation, hence no points of intersection.

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- llafferLv 77 months ago
Let's change those to y's:

y = 3x² - 3x + 1 and y = 2x² - 4x - 3

Both are equal to y so we can set them both equal to each other:

3x² - 3x + 1 = 2x² - 4x - 3

Now simplify to a quadratic and solve:

x² + x + 4 = 0

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -1 ± √(1² - 4(1)(4))] / (2 * 1)

x = [ -1 ± √(1 - 16)] / 2

x = [ -1 ± √(-15)] / 2

x = [ -1 ± √(15)i] / 2

There are no real roots to this quadratic, so the two curves never intersect.

- 7 months ago
3X^2-3X+1= 2X^2-4X-3. X^2+X+2=0.

Then it's (x+1)^2=0.

So X=-1. Then you replace X by -1 in either equations to get the value of Y.