a) \( x+\sqrt{x}=30 \)

To solve the equation \( x + \sqrt{x} = 30 \), let’s introduce a substitution: set \( y = \sqrt{x} \). Then, we can rewrite the equation as \( y^2 + y = 30 \) or \( y^2 + y - 30 = 0 \). This is a quadratic equation that we can solve using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting \( a = 1, b = 1, c = -30 \) into the formula gives us: \[ y = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -30}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 120}}{2} = \frac{-1 \pm 11}{2}. \] This yields two potential values for \( y \), leading us to \( y = 5 \) (since \( y = -6 \) is not valid as a square root). Therefore, \( \sqrt{x} = 5 \) implies \( x = 25 \). Now, let’s check our solution: Substituting \( x = 25 \) back into the original equation verifies that \( 25 + 5 = 30 \), confirming \( x = 25 \) is indeed correct. Moving forward, keep in mind that square roots can introduce extraneous solutions, so it’s always best to double-check your results! Happy solving!