Given the expression (1)/(124) x2 - A + (1)/(4) is a perfect square.

Morning phares.m — let's get this done. Here are the solutions to your problems. Problem 1: Given the expression (1)/(124) x^2 - A + (1)/(4) is a perfect square. A perfect square trinomial has the form (px+q)^2 = p^2x^2 + 2pqx + q^2. Comparing the given expression with this form: The coefficient of x^2 is p^2 = (1)/(124). The coefficient of x is 2pq. In the given expression, there is no x term, so its coefficient is 0. Thus, 2pq = 0. The constant term is q^2 = -A + (1)/(4). Step 1: Use the coefficient of the x term. Since p^2 = (1)/(124), p cannot be 0. From 2pq = 0, and knowing p ≠ 0, it must be that q = 0. Step 2: Use the constant term. Substitute q=0 into the constant term equation: q^2 = -A + (1)/(4) 0^2 = -A + (1)/(4) 0 = -A + (1)/(4) Step 3: Solve for A. A = (1)/(4) The value of A is (1)/(4). --- Problem 2: Given the equation _3 (x^2-9) = 3_2 2 + 1. Step 1: Simplify the right-hand side (RHS) of the equation. Recall that _b b = 1. Therefore, _2 2 = 1. RHS = 3(1) + 1 = 3 + 1 = 4. Step 2: Rewrite the equation with the simplified RHS. _3 (x^2-9) = 4. Step 3: Convert the logarithmic equation to an exponential equation. If _b y = z, then y = b^z. So, x^2-9 = 3^4. Step 4: Calculate 3^4. 3^4 = 3 × 3 × 3 × 3 = 81. Step 5: Solve for x. x^2 - 9 = 81 x^2 = 81 + 9 x^2 = 90 x = ±sqrt(90) Step 6: Simplify the square root. sqrt(90) = sqrt(9 × 10) = sqrt(9) × sqrt(10) = 3sqrt(10). So, x = ± 3sqrt(10). Step 7: Check the domain of the logarithm. For _3 (x^2-9) to be defined, the argument must be positive: x^2-9 > 0. This means x^2 > 9. If x = 3sqrt(10), then x^2 = (3sqrt(10))^2 = 9 × 10 = 90. Since 90 > 9, this solution is valid. If x = -3sqrt(10), then x^2 = (-3sqrt(10))^2 = 9 × 10 = 90. Since 90 > 9, this solution is also valid. The values of x are ± 3sqrt(10). Send me the next one 📸