Question

Rationalize the denominator in 1/√3-√2+1

Answer 1

Answer:

$\sf \dfrac{2 + \sqrt{6} - \sqrt{2}}{4}$

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Step-by-step explanation:

In order to rationalize the denominator, we'll have to multiply both the numerator and the denominator of the fraction by the conjugate of the denominator.

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[Conjugating an expression with two terms basically changes the sign between the two terms from a + to -, or vice-versa.]

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But we've got three terms in the denominator, so we'll group two terms as one, and let the other be a separate term.

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$\dashrightarrow \ \sf \dfrac{1}{\sqrt{3} - \sqrt{2} + 1}$

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On grouping (√3 - √2) we get;

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$\dashrightarrow \ \sf \dfrac{1}{\Big\{\sqrt{3} - \sqrt{2} \Big\}+ 1}$

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Now, the conjugate of the denominator becomes (√3 - √2) - 1, we'll multiply it with the numerator and denominator.

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$\dashrightarrow \ \sf \dfrac{1}{\Big\{ \sqrt{3} - \sqrt{2} \Big\}+ 1} \ \times \ \dfrac{\Big\{\sqrt{3} - \sqrt{2} \Big\} - 1}{\Big\{ \sqrt{3} - \sqrt{2} \Big\} - 1}$

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$\dashrightarrow \ \sf \dfrac{\Big\{\sqrt{3} - \sqrt{2} \Big\} - 1}{\Big\{\Big\{\sqrt{3} - \sqrt{2} \Big\} + 1\Big\} \times \Big\{\Big\{ \sqrt{3} - \sqrt{2} \Big\} - 1\Big\}}$

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Using (a - b)(a - b) = (a - b)² we get;

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$\dashrightarrow \ \sf \dfrac{\Big\{\sqrt{3} - \sqrt{2} \Big\} - 1}{\Big\{\sqrt{3} - \sqrt{2} \Big\}^2 - \Big\{1\Big\}^2}$

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$\dashrightarrow \ \sf \dfrac{\Big\{\sqrt{3} - \sqrt{2} \Big\} - 1}{\Big\{\sqrt{3}\Big\}^2 + \Big\{\sqrt{2}\Big\}^2 - 2\Big\{\sqrt{3}\Big\}\Big\{\sqrt{2}\Big\}- 1}$

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$\dashrightarrow \ \sf \dfrac{\sqrt{3} - \sqrt{2} - 1}{3 + 2 - 2\sqrt{6} - 1}$

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$\dashrightarrow \ \sf \dfrac{\sqrt{3} - \sqrt{2} - 1}{5 - 2\sqrt{6} - 1}$

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$\dashrightarrow \ \sf \dfrac{\sqrt{3} - \sqrt{2} - 1}{4 - 2\sqrt{6}}$

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Now we'll rationalize the denominator once again by mutliplying the numerator and denominator by the conjugate of the denominator.

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$\dashrightarrow \ \sf \dfrac{\sqrt{3} - \sqrt{2} - 1}{4 - 2\sqrt{6}} \times \dfrac{4 + 2\sqrt{6}}{4 + 2\sqrt{6}}$

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$\dashrightarrow \ \sf \dfrac{\Big\{\sqrt{3} - \sqrt{2} - 1\Big\}\Big\{4 + 2\sqrt{6}\Big\}}{\Big\{4 - 2\sqrt{6}\Big\}\Big\{4 + 2\sqrt{6}\Big\}}$

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Using (a - b)(a + b) = a² - b² we get;

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$\dashrightarrow \ \sf \dfrac{\Big\{\sqrt{3}\Big\}\Big\{4\Big\} + \Big\{\sqrt{3}\Big\}\Big\{2\sqrt{6}\Big\} - \Big\{\sqrt{2}\Big\}\Big\{4\Big\} - \Big\{\sqrt{2}\Big\}\Big\{2\sqrt{6}\Big\} - 4 - 2\sqrt{6}}{\Big\{4\Big\}^2 - \Big\{2\sqrt{6}\Big\}^2}$

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$\dashrightarrow \ \sf \dfrac{4\sqrt{3} + 2\sqrt{18}- 4\sqrt{2}- 2\sqrt{12} - 4 - 2\sqrt{6}}{16 - 24}$

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$\dashrightarrow \ \sf \dfrac{4\sqrt{3} + 2\sqrt{2 \times 3 \times 3}- 4\sqrt{2}- 2\sqrt{2 \times 2 \times 3} - 4 - 2\sqrt{6}}{-8}$

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$\dashrightarrow \ \sf \dfrac{4\sqrt{3} + 6\sqrt{2}- 4\sqrt{2}- 4\sqrt{3} - 4 - 2\sqrt{6}}{-8}$

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$\dashrightarrow \ \sf \dfrac{6\sqrt{2}- 4\sqrt{2} - 4 - 2\sqrt{6}}{-8}$

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$\dashrightarrow \ \sf \dfrac{2\sqrt{2} - 4 - 2\sqrt{6}}{-8}$

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$\dashrightarrow \ \sf \dfrac{2\bigg\{\sqrt{2} - 2 - \sqrt{6}\bigg\}}{-8}$

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$\dashrightarrow \ \sf \dfrac{\sqrt{2} - 2 - \sqrt{6}}{-4}$

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$\dashrightarrow \ \sf \dfrac{-\Big\{-\sqrt{2} + 2 + \sqrt{6}\Big\}}{-4}$

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$\dashrightarrow \ \sf \dfrac{-\sqrt{2} + 2 + \sqrt{6}}{4}$

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$\dashrightarrow \ \sf \dfrac{2 + \sqrt{6} - \sqrt{2}}{4}$

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We won't be able to rationalize the fraction further, Hence solved.

Answer 2

$➝ AB = √7x cm.</p><p> </p><p>Math \\ </p><p>5 points</p><p>Now, value of cosA is;</p><p>  \\ </p><p>\dashrightarrow \sf \ cosA \\ = \dfrac{Side \ adjacent \ to \ A}{Hypotenuse}</p><p> </p><p>\dashrightarrow \sf \ cosA = \dfrac{AB}{AC}</p><p>  \\ </p><p>\dashrightarrow \sf \ cosA = \dfrac{\sqrt{7}x}{4x} \\ </p><p> </p><p>\dashrightarrow \sf \ cosA = \bold{\dfrac{\sqrt{7}}{4}}</p><p>$