What are the x-coordinates of the solutions to the system of equations represented by x² + y² = 100 and y = x² ?

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To find the x-coordinates of the solutions to the given system of equations, we need to solve the equations simultaneously. The equations are:

  • Equation 1: x² + y² = 100
  • Equation 2: y = x²

Substituting Equation 2 into Equation 1, we get:

x² + (x²)² = 100

This simplifies to:

x² + x^4 = 100

Rearranging gives:

x^4 + x² - 100 = 0

Now, let’s introduce a substitution to make this easier to solve. Letting u = x², we rewrite the equation as:

u² + u - 100 = 0

This is a standard quadratic equation. We can solve it using the quadratic formula:

u = \frac{-b \pm \sqrt{b² - 4ac}}{2a}
where, a = 1, b = 1, c = -100

Substituting the values gives:

u = \frac{-1 \pm \sqrt{1 + 400}}{2}
= \frac{-1 \pm \sqrt{401}}{2}

This leads to two possible values for u:

  • u₁ = \frac{-1 + \sqrt{401}}{2}
  • u₂ = \frac{-1 – \sqrt{401}}{2} (not valid, as u must be non-negative)

Now, we take the valid solution:

u = \frac{-1 + \sqrt{401}}{2}

Since u = x², we find x by taking the square root:

x = \pm \sqrt{\frac{-1 + \sqrt{401}}{2}}

Thus, the x-coordinates of the solutions to the system of equations are:

  • x₁ = \sqrt{\frac{-1 + \sqrt{401}}{2}}
  • x₂ = -\sqrt{\frac{-1 + \sqrt{401}}{2}}

These x-coordinates correspond to the points where the curves described by the equations intersect.

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