The two numbers are \( a \) and \( b \). Here \( a>b \). The Sum of two numbers is 12 and the product is 32 . A. Multiply with the help of formulae : \( (2 x+3)(2 x-7) \) B. Find out the value of \( 2 a^{2}+2 b^{2} \). C. Prove that, \( (a+2 b)^{2}-5 b^{2}=176 \)

Let's solve each part of the problem step-by-step. --- ### **Given:** - Two numbers \( a \) and \( b \) where \( a > b \). - **Sum:** \( a + b = 12 \) - **Product:** \( ab = 32 \) --- ### **A. Multiply using Formulae: \( (2x + 3)(2x - 7) \)** To multiply the expressions \( (2x + 3) \) and \( (2x - 7) \), we can use the **FOIL (First, Outer, Inner, Last)** method: \[ \begin{align*} (2x + 3)(2x - 7) &= 2x \cdot 2x + 2x \cdot (-7) + 3 \cdot 2x + 3 \cdot (-7) \\ &= 4x^2 - 14x + 6x - 21 \\ &= 4x^2 - 8x - 21 \end{align*} \] **Answer:** \[ (2x + 3)(2x - 7) = 4x^2 - 8x - 21 \] --- ### **B. Find the Value of \( 2a^{2} + 2b^{2} \)** First, let's find \( a^2 + b^2 \): 1. **Square the sum of \( a \) and \( b \):** \[ (a + b)^2 = a^2 + 2ab + b^2 \] Given \( a + b = 12 \) and \( ab = 32 \): \[ 12^2 = a^2 + 2(32) + b^2 \\ 144 = a^2 + 64 + b^2 \\ a^2 + b^2 = 144 - 64 = 80 \] 2. **Compute \( 2a^2 + 2b^2 \):** \[ 2a^2 + 2b^2 = 2(a^2 + b^2) = 2 \times 80 = 160 \] **Answer:** \[ 2a^{2} + 2b^{2} = 160 \] --- ### **C. Prove that \( (a + 2b)^{2} - 5b^{2} = 176 \)** Let's expand and simplify the expression: 1. **Expand \( (a + 2b)^2 \):** \[ (a + 2b)^2 = a^2 + 4ab + 4b^2 \] 2. **Subtract \( 5b^2 \):** \[ (a + 2b)^2 - 5b^2 = a^2 + 4ab + 4b^2 - 5b^2 = a^2 + 4ab - b^2 \] 3. **Substitute known values:** - From part B, \( a^2 + b^2 = 80 \) ⇒ \( a^2 = 80 - b^2 \) - Substitute \( a^2 \) and \( ab \): \[ a^2 + 4ab - b^2 = (80 - b^2) + 4(32) - b^2 = 80 - b^2 + 128 - b^2 = 208 - 2b^2 \] 4. **Find \( b^2 \):** - Use the product \( ab = 32 \) and the sum \( a + b = 12 \). - Solve for \( b \): \[ a = 12 - b \\ (12 - b)b = 32 \\ 12b - b^2 = 32 \\ b^2 - 12b + 32 = 0 \\ \] - Solve the quadratic equation: \[ b = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2} = \frac{12 \pm 4}{2} \] \[ b = 8 \ \text{or} \ b = 4 \] - Since \( a > b \), \( b = 4 \) and \( a = 8 \). 5. **Calculate \( 208 - 2b^2 \):** \[ 208 - 2(4)^2 = 208 - 2(16) = 208 - 32 = 176 \] **Thus,** \[ (a + 2b)^{2} - 5b^{2} = 176 \] **Answer:** \[ (a + 2b)^{2} - 5b^{2} = 176 \] --- ### **Summary of Answers:** - **A.** \( 4x^2 - 8x - 21 \) - **B.** \( 160 \) - **C.** \( 176 \)