Number system practice

Operation on real numbers & algebraic identities

Question: Simplify the following:

(a) \({8 + \sqrt{5})}) \({8 - \sqrt{5})})

(b) \({10 + \sqrt{3})}) \({6 + \sqrt{2})})

(c) \{(\sqrt {3} + \sqrt {11})}^2) + \{(\sqrt {3} - \sqrt {11})}^2)

Video Explanation

Explanatory Answer

(a) \({8 + \sqrt{5})}) \({8 - \sqrt{5})})

The expression is of the form (x + y) (x – y) = x2 – y2

So, \({8 + \sqrt{5})}) \({8 - \sqrt{5})} = ({8^2} - \sqrt{5}^2))
= 64 – 5 = 59

(b) \({10 + \sqrt{3})}) \({6 + \sqrt{2})})

Simplifying in such terms is the same as expanding the terms of the expression
= \{(10 \times 6)} + 10 \times \sqrt{2} + \sqrt{3} \times 6 + \sqrt{2} \times \sqrt{3})
= \60 + 10\sqrt{2} + 6\sqrt{3} + \sqrt{6})

(c) \{(\sqrt {3} + \sqrt {11})}^2 + {(\sqrt {3} - \sqrt {11})}^2)

\{(\sqrt {3} + \sqrt {11})}^2) = \{(\sqrt {3})}^2) + \{(\sqrt {11})}^2 + 2 \times \sqrt{3} \times \sqrt{11})

\= 3 + 11 + 2\sqrt{33} = 14 + 2\sqrt{33})

\{(\sqrt {3} - \sqrt {11})}^2) = \{(\sqrt {3})}^2) + \{(\sqrt {11})}^2 - 2 \times \sqrt{3} \times \sqrt{11})

\= 3 + 11 - 2\sqrt{33} = 14 - 2\sqrt{33})

Thus, \{(\sqrt {3} + \sqrt {11})}^2) + \{(\sqrt {3} - \sqrt {11})}^2 = 14 + 2\sqrt{33} + 14 - 2\sqrt{33})
= 28