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▪️Grade - 9
▪️Chapter - Polynomials
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Answer 1

Question :

Rationalise the denominator of $\tt \frac{1}{7+3\sqrt{2}}$

Solution :

$\tt \frac{1}{7+3\sqrt{2}}$

To rationalise any denominator we have to multiply the denominator with his conjugate.

$\tt \implies \frac{1(7-3\sqrt{2}}{(7+3\sqrt{2})(7-3\sqrt{2})}$

$\tt \implies \frac{7-3\sqrt{2}}{{7}^{2}-{(3\sqrt{2})}^{2}}$

$\tt \implies \frac{7-3\sqrt{2}}{49-18}$

$\tt \implies \frac{7-3\sqrt{2}}{31}$

Now the denominator is rationalised.

Answer 2

Answer :

$\sf \to \dfrac{7 - 3 \sqrt{2} }{31}$

Explanation :

Given fraction,

$\rm \to \dfrac{1}{7 + 3\sqrt{2} }$

Rationalising the denominator.

Multiply the fraction by the conjugate of the denominator i.e., 7 + 3√2

$\rm \to \: \dfrac{1}{7 + 3 \sqrt{2} } \times \dfrac{ \: {7} - 3 \sqrt{2}}{7 - 3 \sqrt{2} }$

$\rm \to \: \dfrac{1(7 - 3 \sqrt{2}) }{(7 {)}^{2} + (3 \sqrt{2} {)}^{2} } \: \: \: \{ \because \: (a - b)(a + b) = {a}^{2} - {b}^{2} \}$

$\rm \to \: \dfrac{7 - 3 \sqrt{2} }{49 - 18}$

$\rm \to \: \dfrac{7 - 3 \sqrt{2} }{31}$

Required answer : $\sf \to \dfrac{7 - 3 \sqrt{2} }{31}$