To solve the system of equations given by:
- \( y = 2x^2 - 9 \)
- \( y = x^2 - 10x - 25 \)
we can set the two equations equal to each other since they both equal \( y \):
\[ 2x^2 - 9 = x^2 - 10x - 25 \]
Now, let's move all terms to one side of the equation:
\[ 2x^2 - x^2 + 10x - 25 + 9 = 0 \] \[ x^2 + 10x - 16 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 10 \), and \( c = -16 \).
Calculating the discriminant:
\[ b^2 - 4ac = 10^2 - 4(1)(-16) = 100 + 64 = 164 \]
Now we can plug in the values into the quadratic formula:
\[ x = \frac{-10 \pm \sqrt{164}}{2(1)} = \frac{-10 \pm \sqrt{164}}{2} \]
Simplifying \( \sqrt{164} \):
\[ \sqrt{164} = \sqrt{4 \cdot 41} = 2\sqrt{41} \]
So, we have:
\[ x = \frac{-10 \pm 2\sqrt{41}}{2} = -5 \pm \sqrt{41} \]
This gives us the two solutions for \( x \):
\[ x_1 = -5 + \sqrt{41} \] \[ x_2 = -5 - \sqrt{41} \]
Now we calculate the corresponding \( y \) values by substituting back into one of the original equations, for example \( y = 2x^2 - 9 \).
For \( x_1 = -5 + \sqrt{41} \):
\[ y_1 = 2(-5 + \sqrt{41})^2 - 9 \] Calculating \( (-5 + \sqrt{41})^2 \):
\[ (-5 + \sqrt{41})^2 = 25 - 10\sqrt{41} + 41 = 66 - 10\sqrt{41} \] Thus,
\[ y_1 = 2(66 - 10\sqrt{41}) - 9 = 132 - 20\sqrt{41} - 9 = 123 - 20\sqrt{41} \]
For \( x_2 = -5 - \sqrt{41} \):
\[ y_2 = 2(-5 - \sqrt{41})^2 - 9 \] Calculating \( (-5 - \sqrt{41})^2 \):
\[ (-5 - \sqrt{41})^2 = 25 + 10\sqrt{41} + 41 = 66 + 10\sqrt{41} \] Thus,
\[ y_2 = 2(66 + 10\sqrt{41}) - 9 = 132 + 20\sqrt{41} - 9 = 123 + 20\sqrt{41} \]
Now we have our solutions:
- \( \left(-5 - \sqrt{41}, 123 + 20\sqrt{41}\right) \)
- \( \left(-5 + \sqrt{41}, 123 - 20\sqrt{41}\right) \)
Therefore, the solutions in order of least \( x \)-value first:
\[ \left(-5 - \sqrt{41}, 123 + 20\sqrt{41}\right) \quad \text{and} \quad \left(-5 + \sqrt{41}, 123 - 20\sqrt{41}\right) \]