Hailey claims that the equation \( (x-3)^{2}+(x+5)^{2}=0 \) has no real solutions. Justify with working if you agree with her.

To determine if Hailey's claim is correct, we need to solve the equation \( (x-3)^{2}+(x+5)^{2}=0 \) and see if there are any real solutions. Let's start by expanding the squares: \[ (x-3)^{2} = x^2 - 6x + 9 \] \[ (x+5)^{2} = x^2 + 10x + 25 \] Now, we add these two expressions together: \[ x^2 - 6x + 9 + x^2 + 10x + 25 = 0 \] Combine like terms: \[ 2x^2 + 4x + 34 = 0 \] Now we have a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where \( a = 2 \), \( b = 4 \), and \( c = 34 \). To determine if this quadratic equation has real solutions, we can use the discriminant, which is given by the formula \( \Delta = b^2 - 4ac \). Let's calculate the discriminant: \[ \Delta = 4^2 - 4 \cdot 2 \cdot 34 \] \[ \Delta = 16 - 272 \] \[ \Delta = -256 \] Since the discriminant \( \Delta \) is negative, the quadratic equation \( 2x^2 + 4x + 34 = 0 \) has no real solutions. This means that Hailey's claim is correct, and the equation \( (x-3)^{2}+(x+5)^{2}=0 \) indeed has no real solutions.