Question

rationalize the denominator
$10 \div 2 \sqrt{2 } + \sqrt{3}$
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Answer 1

Answer:

In elementary algebra, root rationalisation is a process by which radicals in the denominator of an algebraic fraction are eliminated.

If the denominator is a monomial in some radical, say {\displaystyle a{\sqrt[{n}]{x}}^{k},} {\displaystyle a{\sqrt[{n}]{x}}^{k},} with k < n, rationalisation consists of multiplying the numerator and the denominator by {\displaystyle {\sqrt[{n}]{x}}^{n-k},} {\displaystyle {\sqrt[{n}]{x}}^{n-k},} and replacing {\displaystyle {\sqrt[{n}]{x}}^{n}} {\displaystyle {\sqrt[{n}]{x}}^{n}} by {\displaystyle \left|x\right|,} {\displaystyle \left|x\right|,} if n is even or by x if n is odd (if k ≥ n, the same replacement allows us to reduce k until it becomes lower than n.

If the denominator is linear in some square root, say {\displaystyle a+b{\sqrt {x}},} {\displaystyle a+b{\sqrt {x}},} rationalisation consists of multiplying the numerator and the denominator by {\displaystyle a-b{\sqrt {x}},} {\displaystyle a-b{\sqrt {x}},} and expanding the product in the denominator.

This technique may be extended to any algebraic denominator, by multiplying the numerator and the denominator by all algebraic conjugates of the denominator, and expanding the new denominator into the norm of the old denominator. However, except in special cases, the resulting fractions may have huge numerators and denominators, and, therefore, the technique is generally used only in the above elementary cases.

Answer 2

Answer:

$\huge\underbrace\mathfrak\red {Answer}$

10/2√2+√3=>10/2√2+√3*(2√2-√3)/(2√2-√3)

=10(2√2-√3)/(2√2)^2-(√3)^2

=10(2√2-√3)/(8-3)

=10(2√2-√3)/5

=2(2√2-√3)

=4√2-2√3

Hope This Helps You

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