Mathematics: Hi:) anyone able to help with 4(a) ? question in the pic attached :) thanks!

Sin² x + sin x cos x = 0sin x ( sin x + cos x ) = 0sin x = 0 ; x 1 = k π ; k ∈ Zsin x + cos x = 0 / ²sin² x + 2 sin x cos x + cos² x = 02 sin x cos x + 1 =0sin ( 2 x ) = - 1x = 3π / 2 + 2 kπ / : 2x 2 = 3π / 4 + kπ ; k ∈ Z ...

Hi:) anyone able to help with 4(a) ? question in

Ответ:

lovebug3563

21.01.2022

Mathematics

Answers:a = 2b = -1

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Work Shown:

$x\sqrt{24}+\sqrt{96} = \sqrt{108}+x\sqrt{12}\\\\x\sqrt{24}-x\sqrt{12} = \sqrt{108}-\sqrt{96}\\\\x(\sqrt{24}-\sqrt{12}) = \sqrt{108}-\sqrt{96}\\\\x = \frac{\sqrt{108}-\sqrt{96}}{\sqrt{24}-\sqrt{12}}\\\\x = \frac{6\sqrt{3}-4\sqrt{6}}{2\sqrt{6}-2\sqrt{3}}\\\\$

$x = \frac{3\sqrt{3}-2\sqrt{6}}{\sqrt{6}-\sqrt{3}}\\\\x = \frac{(3\sqrt{3}-2\sqrt{6})(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})} \text{ rationalizing denominator}\\\\x = \frac{3\sqrt{2}-3}{(\sqrt{6})^2-(\sqrt{3})^2} \text{ see note below; see image attachment below}\\\\x = \frac{3(\sqrt{2}-1)}{6-3}\\\\x = \frac{3(\sqrt{2}-1)}{3}\\\\x = \sqrt{2}-1\\\\$

This is in the form $\sqrt{a}+b$ with a = 2 and b = -1

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note: for this line, I expanded each pair of multiplying binomials. In the numerator, I used the box method as shown in the diagram below. Each inner cell is the result of multiplying the corresponding outer cell expressions. Example: for row2, column1, we have $3\sqrt{3} * \sqrt{3} = 3\sqrt{3*3} = 3\sqrt{9} = 3*3 = 9$ . The other cells are filled out in a similar fashion. In the denominator, I used the difference of squares rule.

Hi:) anyone able to help with 4(a) ? question in the pic attached :) thanks!

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Hi:) anyone able to help with 4(a) ? question in the pic attached :) thanks!

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