Math, asked by sagar766, 1 year ago

3-2×√2=a+b×√2then b =​

Answers

Answered by AbhijithPrakash
5

Answer:

3-2\sqrt{2}=a+b\sqrt{2}\quad :\quad b=\dfrac{3\sqrt{2}-4-\sqrt{2}a}{2}

Step-by-step explanation:

\rule{300}{1.05}

3-2\sqrt{2}=a+b\sqrt{2}

\rule{300}{1.05}

\mathrm{Switch\:sides}

a+b\sqrt{2}=3-2\sqrt{2}

\rule{300}{1.05}

\mathrm{Subtract\:}a\mathrm{\:from\:both\:sides}

a+b\sqrt{2}-a=3-2\sqrt{2}-a

\rule{300}{1.05}

\mathrm{Simplify}

b\sqrt{2}=3-2\sqrt{2}-a

\rule{300}{1.05}

\mathrm{Divide\:both\:sides\:by\:}\sqrt{2}

\dfrac{b\sqrt{2}}{\sqrt{2}}=\dfrac{3}{\sqrt{2}}-\dfrac{2\sqrt{2}}{\sqrt{2}}-\dfrac{a}{\sqrt{2}}

\rule{300}{1.05}

\mathrm{Simplify}

\rule{300}{0.5}

\mathrm{Simplify\:}\dfrac{b\sqrt{2}}{\sqrt{2}}:\quad b

\rule{300}{0.5}

\rule{300}{0.5}

\mathrm{Simplify\:}\dfrac{3}{\sqrt{2}}-\dfrac{2\sqrt{2}}{\sqrt{2}}-\dfrac{a}{\sqrt{2}}

\rule{300}{0.5}

\dfrac{3}{\sqrt{2}}-\dfrac{2\sqrt{2}}{\sqrt{2}}-\dfrac{a}{\sqrt{2}}

\rule{300}{0.5}

\mathrm{Apply\:rule}\:\dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}

=\dfrac{3-2\sqrt{2}-a}{\sqrt{2}}

\rule{300}{0.5}

\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{\sqrt{2}}{\sqrt{2}}

=\dfrac{\left(3-2\sqrt{2}-a\right)\sqrt{2}}{\sqrt{2}\sqrt{2}}

\rule{300}{1.05}

\mathrm{Simplify}

=\dfrac{3\sqrt{2}-4-\sqrt{2}a}{2}

\rule{300}{1.05}

b=\dfrac{3\sqrt{2}-4-\sqrt{2}a}{2}

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