May 2026

Simplify (3 + √3)(5 − √2) 🎥 Video Solution: 📘 Simplify: \[ (3 + \sqrt{3})(5 – \sqrt{2}) \] ✏️ Solution: \[ = 3\cdot5 – 3\sqrt{2} + 5\sqrt{3} – \sqrt{3}\sqrt{2} \] \[ = 15 – 3\sqrt{2} + 5\sqrt{3} – \sqrt{6} \] ✅ Final Answer: \(15 – 3\sqrt{2} + 5\sqrt{3} – \sqrt{6}\) Next Question / Full Exercise

Simplify (4 + √7)(3 + √2) 🎥 Video Solution: 📘 Simplify: \[ (4 + \sqrt{7})(3 + \sqrt{2}) \] ✏️ Solution: \[ = 4 \cdot 3 + 4\sqrt{2} + 3\sqrt{7} + \sqrt{7}\sqrt{2} \] \[ = 12 + 4\sqrt{2} + 3\sqrt{7} + \sqrt{14} \] ✅ Final Answer: \(12 + 4\sqrt{2} + 3\sqrt{7} + \sqrt{14}\) Next Question /

Simplify 4√28 ÷ 3√7 🎥 Video Solution: 📘 Simplify: \[ \frac{4\sqrt{28}}{3\sqrt{7}} \] ✏️ Solution: \[ = \frac{4}{3} \times \sqrt{\frac{28}{7}} \] \[ = \frac{4}{3} \times \sqrt{4} \] \[ = \frac{4}{3} \times 2 \] \[ = \frac{8}{3} \] ✅ Final Answer: \(\frac{8}{3}\) Next Question / Full Exercise

Simplify ⁴√1250 ÷ ⁴√2 🎥 Video Solution: 📘 Simplify: \[ \frac{\sqrt[4]{1250}}{\sqrt[4]{2}} \] ✏️ Solution: \[ \frac{\sqrt[4]{1250}}{\sqrt[4]{2}} = \sqrt[4]{\frac{1250}{2}} \] \[ = \sqrt[4]{625} \] \[ = \sqrt[4]{5^4} \] \[ = 5 \] ✅ Final Answer: \(5\) Next Question / Full Exercise

Simplify ∛4 × ∛16 🎥 Video Solution: 📘 Simplify: \[ \sqrt[3]{4} \times \sqrt[3]{16} \] ✏️ Solution: \[ \sqrt[3]{4} \times \sqrt[3]{16} = \sqrt[3]{4 \times 16} \] \[ = \sqrt[3]{64} \] \[ = 4 \] ✅ Final Answer: \(4\) Next Question / Full Exercise

Assertion Reason Exponents 🎥 Watch Video Solution Q. Assertion–Reason Type Question Statement-1: \( \left(\sqrt[m]{\sqrt[n]{a}}\right)^{mn} = a \) Statement-2: \( ((a^m)^n)^p = a^{mnp} \) ✏️ Solution \( \sqrt[n]{a} = a^{1/n} \) \( \sqrt[m]{\sqrt[n]{a}} = (a^{1/n})^{1/m} = a^{1/(mn)} \) \( \left(a^{1/(mn)}\right)^{mn} = a \) So Statement-1 is TRUE Statement-2 is also TRUE (law of exponents) Statement-2 correctly

Assertion Reason Infinite Roots 🎥 Watch Video Solution Q. Assertion–Reason Type Question Statement-1: \( \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} = 3 \) Statement-2: \( \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} = x,\ x>0 \) ✏️ Solution Let \( y = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} \) Then \( y = \sqrt{6+y} \) Squaring: \( y^2 = 6 + y \) \( y^2 – y – 6 = 0

Assertion Reason Infinite Roots 🎥 Watch Video Solution Q. Assertion–Reason Type Question Statement-1: \( \sqrt{5\sqrt{5\sqrt{5\cdots}}} = 5\sqrt{5} \) Statement-2: \( \sqrt{x\sqrt{x\sqrt{x\cdots}}} = x,\ x>0 \) ✏️ Solution Let \( y = \sqrt{5\sqrt{5\sqrt{5\cdots}}} \) Then \( y = \sqrt{5y} \) Squaring: \( y^2 = 5y \) \( y(y-5) = 0 \Rightarrow y = 5 \) (since

Assertion Reason Nested Roots 🎥 Watch Video Solution Q. Assertion–Reason Type Question Statement-1: \( \sqrt{7\sqrt{7\sqrt{7\sqrt{7}}}} = \sqrt[16]{7^{15}} \) Statement-2: \( \sqrt{a\sqrt{a\sqrt{a\cdots}}} \ (n\text{ terms}) = a^{\frac{2^n – 1}{2^n}} \) ✏️ Solution Number of nested roots = 4 Using formula: \( = 7^{\frac{2^4 – 1}{2^4}} = 7^{\frac{16 – 1}{16}} = 7^{15/16} \) \( = \sqrt[16]{7^{15}} \)

Assertion Reason Exponents 🎥 Watch Video Solution Q. Assertion–Reason Type Question Statement-1: If \( a^x = b^y = c^z = abc \), then \( xy + yz + zx + xyz = 1 \) Statement-2: If \( a^n = k \), then \( a = k^{1/n} \) ✏️ Solution Let \( a^x = b^y =