Q.3 \( \sqrt{13^{2}+28 \div 4-(3)^{3}+107}=(2)^{2} \) \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } 16 & \text { (c) } 256 \quad \text { (d) } 4 & \text { (e) }(256)^{2} \\ \text { Q.- } 51 a^{2}+b^{2} & =234 \quad \text { \& } a b=108 \quad \text { find value } \\ \frac{a+b}{a-b}\end{array} \)

Let's solve the first part of the problem step by step: Given: \( \sqrt{13^{2}+28 \div 4-(3)^{3}+107}=(2)^{2} \) 1. Simplify the expression inside the square root: \( 13^{2} = 169 \) \( 28 \div 4 = 7 \) \( (3)^{3} = 27 \) \( 169 + 7 - 27 + 107 = 156 \) 2. Substitute the simplified expression back into the equation: \( \sqrt{156} = (2)^{2} \) 3. Calculate the square root of 156: \( \sqrt{156} \approx 12.49 \) 4. Compare the result with \( (2)^{2} \): \( 12.49 \neq 4 \) Therefore, the correct answer for the first part is not provided in the options given. Now, let's move on to the second part of the problem: Given: \( 51a^{2}+b^{2} = 234 \) and \( ab = 108 \) We need to find the value of \( \frac{a+b}{a-b} \). 1. We can rewrite the equation \( 51a^{2}+b^{2} = 234 \) as \( 51a^{2} = 234 - b^{2} \). 2. Substitute \( ab = 108 \) into the equation: \( 51a^{2} = 234 - b^{2} \) \( 51a^{2} = 234 - \frac{108^{2}}{a^{2}} \) 3. Solve for \( a \) and \( b \) using the given equations. 4. Calculate \( \frac{a+b}{a-b} \) using the values of \( a \) and \( b \) obtained in step 3. Let's proceed with solving the second part of the problem. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}51a^{2}+b^{2}=234\\ab=108\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}51a^{2}+b^{2}=234\\a=\frac{108}{b}\end{array}\right.\) - step2: Substitute the value of \(a:\) \(51\left(\frac{108}{b}\right)^{2}+b^{2}=234\) - step3: Simplify: \(\frac{594864+b^{4}}{b^{2}}=234\) - step4: Cross multiply: \(594864+b^{4}=b^{2}\times 234\) - step5: Simplify the equation: \(594864+b^{4}=234b^{2}\) - step6: Move the expression to the left side: \(594864+b^{4}-234b^{2}=0\) - step7: Solve using substitution: \(594864+t^{2}-234t=0\) - step8: Rewrite in standard form: \(t^{2}-234t+594864=0\) - step9: Solve using the quadratic formula: \(t=\frac{234\pm \sqrt{\left(-234\right)^{2}-4\times 594864}}{2}\) - step10: Simplify the expression: \(t=\frac{234\pm \sqrt{-2324700}}{2}\) - step11: The expression is undefined: \(t \notin \mathbb{R}\) - step12: Calculate: \(b \notin \mathbb{R}\) - step13: The system of equations has no solution in the set of real numbers: \(\left(a,b\right) \notin \mathbb{R}^{2}\) - step14: Alternative Form: \(\textrm{No real solution}\) The system of equations \(51a^{2}+b^{2} = 234\) and \(ab = 108\) has no real solution. This means that there are no real values of \(a\) and \(b\) that satisfy both equations simultaneously. Therefore, we cannot find the value of \( \frac{a+b}{a-b} \) based on the given equations.