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4, 9, 25, 49, ….

The given sequence is the square on prime number 2, 3, 5, 7, 11 etc.  The next number is 121.

√10 * √15 =

The given expression is \(\begin{array}{l}\sqrt{10} \times \sqrt{15}\\=\sqrt{5 \times 2} \times \sqrt{5 \times 3}\\=\sqrt{5 \times 5 \times 2 \times 3}\\=5 \sqrt{6}\end{array}... View Article

Find the continued product: (x – 3) (x + 3) (x2 + 9).

Consider the given expression. \(\begin{array}{l}\begin{array}{l} (\mathbf{x}-3)(\mathbf{x}+3)\left(\mathbf{x}^{2}+9\right) \\ (\mathbf{a}+\mathbf{b})(\mathbf{a}-\mathbf{b})=\mathbf{a}^{2}-\mathbf{b}^{2} \\ =\left(\mathbf{x}^{2}-9\right)\left(\mathbf{x}^{2}+9\right) \\ =\mathbf{x}^{4}-\mathbf{8 1} \end{array}\end{array} \)

If a + b = 5 and ab = 6, then find a2 + b2.

To find a2 + b2. \(\begin{array}{l}(\mathbf{a}+\mathbf{b})^{2}=\mathbf{a}^{2}+\mathbf{b}^{2}+\mathbf{2 a b} \\ =>\mathbf{a}^{2}+\mathbf{b}^{2}=(\mathbf{a}+\mathbf{b})^{2}-2 \mathrm{ab} \\ \mathbf{a}+\mathbf{b}=5 \text { and } \mathrm{ab}=6 \\ \mathbf{a}^{2}+\mathbf{b}^{2}=(\mathbf{a}+\mathbf{b})^{2}-2... View Article