Fill in the blanks under each radical to make the statement true. \( (3-\sqrt{\square})(3+\sqrt{ })=4 \)

To solve the equation \( (3-\sqrt{a})(3+\sqrt{b})=4 \), we can use the difference of squares formula, which states that \( (x-y)(x+y) = x^2 - y^2 \). Here, we can identify: - \( x = 3 \) - \( y = \sqrt{b} \) Thus, we can rewrite the expression as: \[ (3-\sqrt{a})(3+\sqrt{b}) = 3^2 - (\sqrt{b})^2 = 9 - b \] Setting this equal to 4 gives us: \[ 9 - b = 4 \] Now, we can solve for \( b \): \[ b = 9 - 4 = 5 \] Next, we can also express the equation in terms of \( a \): \[ (3-\sqrt{a})(3+\sqrt{b}) = 4 \] Substituting \( b = 5 \): \[ (3-\sqrt{a})(3+\sqrt{5}) = 4 \] Now, we can expand this: \[ 3(3+\sqrt{5}) - \sqrt{a}(3+\sqrt{5}) = 4 \] Calculating \( 3(3+\sqrt{5}) \): \[ 9 + 3\sqrt{5} - \sqrt{a}(3+\sqrt{5}) = 4 \] Rearranging gives: \[ 9 + 3\sqrt{5} - 4 = \sqrt{a}(3+\sqrt{5}) \] \[ 5 + 3\sqrt{5} = \sqrt{a}(3+\sqrt{5}) \] To isolate \( \sqrt{a} \), we can divide both sides by \( 3+\sqrt{5} \): \[ \sqrt{a} = \frac{5 + 3\sqrt{5}}{3+\sqrt{5}} \] Now, we can rationalize the denominator: \[ \sqrt{a} = \frac{(5 + 3\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} \] Calculating the denominator: \[ (3+\sqrt{5})(3-\sqrt{5}) = 9 - 5 = 4 \] Calculating the numerator: \[ (5 + 3\sqrt{5})(3-\sqrt{5}) = 15 - 5\sqrt{5} + 9\sqrt{5} - 15 = 4\sqrt{5} \] Thus: \[ \sqrt{a} = \frac{4\sqrt{5}}{4} = \sqrt{5} \] Squaring both sides gives: \[ a = 5 \] In conclusion, the values to fill in the blanks are: - \( a = 5 \) - \( b = 5 \) So the completed statement is: \[ (3-\sqrt{5})(3+\sqrt{5})=4 \]