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Svetlanka [38]
3 years ago
15

(function form)

Mathematics
1 answer:
Mariana [72]3 years ago
3 0
Seven and a half hours
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If a = 6 - \sqrt{35}, find the value of a² + \frac{1}{a²}

<h2><u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

a = 6 - \sqrt{35} (Given)

Now, first of all we have to find the value of \frac{1}{a}.

So, \frac{1}{a} = \frac{1}{6 \: - \: \sqrt{35}}

\frac{1}{a} = \frac{6 \: + \sqrt{35}}{(6 \: - \: \sqrt{35})(6 \: + \: \sqrt{35})}

\frac{1}{a} = \frac{6 \: + \sqrt{35}}{(6)² \: - \: (\sqrt{35})²}

\frac{1}{a} = \frac{6 \: + \sqrt{35}}{36 \: - \: 35} (\because \sqrt{35} \: × \: \sqrt{35} \: = \: 35)

\frac{1}{a} = \frac{6 \: + \: \sqrt{35}}{1}

\frac{1}{a} = 6 + \sqrt{35}

Now, we have to find the value of a² + \frac{1}{a²}

So, a² + \frac{1}{a²} = (a \: + \frac{1}{a})² - 2.a.\frac{1}{a}

a² + \frac{1}{a²} = (6 \: - \sqrt{35} \: + \: 6 \: + \sqrt{35})² - 2

a² + \frac{1}{a²} = (12)² - 2

a² + \frac{1}{a²} = 144 - 2

a² + \frac{1}{a²} = 142

The value of a² + \frac{1}{a²} is <u>1</u><u>4</u><u>2</u>. [Answer]

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3a^2-7a+6= <br> ( for a= -2 )
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Substitute a = - 2 into the expression and evaluate

3(- 2)² - 7(- 2) + 6 = 3(4) + 14 + 6 = 12 + 14 + 6 = 32

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