This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the problems. Problem 3: If and are the roots of the equation 3x^2 - 8x - 4 = 0, find the value of ^2 + ^2. Step 1: Identify the coefficients of the quadratic equation. The given quadratic equation is 3x^2 - 8x - 4 = 0. Comparing this to the standard form ax^2 + bx + c = 0, we have: a = 3 b = -8 c = -4 Step 2: Find the sum and product of the roots. For a quadratic equation ax^2 + bx + c = 0, the sum of the roots ( + ) and the product of the roots () are given by: + = -(b)/(a) = -(-8)/(3) = (8)/(3) = (c)/(a) = (-4)/(3) Step 3: Use the identity for ^2 + ^2. We know that ( + )^2 = ^2 + 2 + ^2. Rearranging this identity to find ^2 + ^2: ^2 + ^2 = ( + )^2 - 2 Step 4: Substitute the values of the sum and product of roots into the identity. ^2 + ^2 = ((8)/(3))^2 - 2((-4)/(3)) ^2 + ^2 = (64)/(9) - (-(8)/(3)) ^2 + ^2 = (64)/(9) + (8)/(3) Step 5: Add the fractions to find the final value. To add the fractions, find a common denominator, which is 9. ^2 + ^2 = (64)/(9) + (8 × 3)/(3 × 3) ^2 + ^2 = (64)/(9) + (24)/(9) ^2 + ^2 = (64 + 24)/(9) ^2 + ^2 = (88)/(9) The final answer is (88)/(9). --- Problem 4: Find the coefficient of x^11 in the expression of (x^2 + (1)/(x))^10. Step 1: Write the general term of the binomial expansion. The binomial theorem states that the general term T_r+1 in the expansion of (a+b)^n is given by: T_r+1 = nr a^n-r b^r In this expression, a = x^2, b = (1)/(x) = x^-1, and n = 10. Substitute these values into the general term formula: T_r+1 = 10r (x^2)^10-r (x^-1)^r Step 2: Simplify the powers of x. T_r+1 = 10r x^2(10-r) x^-r T_r+1 = 10r x^20-2r x^-r Combine the terms with x: T_r+1 = 10r x^20-2r-r T_r+1 = 10r x^20-3r Step 3: Set the exponent of x equal to 11 to find r. We are looking for the coefficient of x^11, so we set the exponent 20-3r equal to 11: 20 - 3r = 11 3r = 20 - 11 3r = 9 r = 3 Step 4: Calculate the binomial coefficient for r=3. The coefficient of x^11 is 10r when r=3. Coefficient = 103 103 = (10!)/(3!(10-3)!) = (10!)/(3!7!) 103 = (10 × 9 × 8 × 7!)/(3 × 2 × 1 × 7!) 103 = (10 × 9 × 8)/(3 × 2 × 1) 103 = (720)/(6) 103 = 120 The final answer is 120. 3 done, 2 left today. You're making progress.