Questions: Graph each equation of the system. Solve the system to find the points of intersection. x^2 + y^2 = 25 y = x^2 - 100

Solution Steps

The first equation is: \[ x^2 + y^2 = 25 \] Solving for \( y \), we get: \[ y^2 = 25 - x^2 \] \[ y = \pm \sqrt{25 - x^2} \]

The second equation is: \[ y = x^2 - 100 \] Substitute \( y = \pm \sqrt{25 - x^2} \) into this equation: \[ \pm \sqrt{25 - x^2} = x^2 - 100 \]

Square both sides to eliminate the square root: \[ 25 - x^2 = (x^2 - 100)^2 \] Expanding the right side: \[ 25 - x^2 = x^4 - 200x^2 + 10000 \] Rearrange the equation: \[ x^4 - 199x^2 + 9975 = 0 \] This is a quadratic in terms of \( x^2 \). Let \( z = x^2 \): \[ z^2 - 199z + 9975 = 0 \] Using the quadratic formula: \[ z = \frac{199 \pm \sqrt{199^2 - 4 \times 9975}}{2} \] \[ z = \frac{199 \pm \sqrt{39601 - 39900}}{2} \] \[ z = \frac{199 \pm \sqrt{701}}{2} \] Calculate the roots: \[ z_1 = \frac{199 + \sqrt{701}}{2} \approx 199.4186 \] \[ z_2 = \frac{199 - \sqrt{701}}{2} \approx 0.5814 \] Since \( z = x^2 \), solve for \( x \): \[ x = \pm \sqrt{199.4186} \quad \text{and} \quad x = \pm \sqrt{0.5814} \]

Final Answer