Fresh day segunolude0375, let's solve. To show that $(\sqrt{5} + \sqrt{13})^2 > 34$, we will expand the left side of the inequality. Step 1: Expand the expression $(\sqrt{5} + \sqrt{13})^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$. Here, $a = \sqrt{5}$ and $b = \sqrt{13}$. $$(\sqrt{5} + \sqrt{13})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{13}) + (\sqrt{13})^2$$ Step 2: Simplify the terms. $$(\sqrt{5})^2 = 5$$ $$(\sqrt{13})^2 = 13$$ $$2(\sqrt{5})(\sqrt{13}) = 2\sqrt{5 \times 13} = 2\sqrt{65}$$ Step 3: Substitute the simplified terms back into the expanded expression. $$(\sqrt{5} + \sqrt{13})^2 = 5 + 2\sqrt{65} + 13$$ $$(\sqrt{5} + \sqrt{13})^2 = 18 + 2\sqrt{65}$$ Step 4: Now we need to show that $18 + 2\sqrt{65} > 34$. Subtract 18 from both sides of the inequality: $$2\sqrt{65} > 34 - 18$$ $$2\sqrt{65} > 16$$ Step 5: Divide both sides by 2. $$\sqrt{65} > \frac{16}{2}$$ $$\sqrt{65} > 8$$ Step 6: Square both sides of the inequality to remove the square root. Since both sides are positive, the inequality direction remains the same. $$(\sqrt{65})^2 > 8^2$$ $$65 > 64$$ Step 7: Conclude the proof. Since $65 > 64$ is a true statement, the original inequality $(\sqrt{5} + \sqrt{13})^2 > 34$ is also true. We have shown that $(\sqrt{5} + \sqrt{13})^2 > 34$. $\boxed{\text{Proven}}$ Send me the next one 📸