Question

$\huge\tt\red{P}\tt\pink{L}\tt\blue{E}\tt\green{A}\tt\purple{S}\tt\orange{E}$
$\huge\tt\red{A}\tt\pink{N}\tt\blue{S}\tt\green{W}\tt\purple{E}\tt\orange{R}$

$\huge\pink \: \mathfrak{Question}$
correct answer will be marked as brainlist
and wrong answer will be reported ​

Answer 1

★ Given:

${⇥\rm{a + \frac{1}{a} = 3 }}$

★ To Find:

${⇥\rm{ {a}^{2} + \frac{1}{ {a}^{2} } }}$

★ Solution:

${\rm{ {(a + \frac{1}{a} )}^{2} }} = {\rm{ {a}^{2} + 2 \times {\cancel a \times \frac{1}{\cancel{a}} + \frac{1}{ {a}^{2} } }}}$

${\rm{( {a + \frac{1}{a}) }^{2} }} = {a}^{2} + \frac{1}{{a}^{2} } + 2 \\ {\rm{( {a + \frac{1}{a} )}^{2} - 2 }} = {a}^{2} + \frac{1}{{a}^{2} } \\ {\rm{ ({3})^{2} - 2 = {a}^{2} + \frac{1}{{a}^{2} }}} \\ {\rm{9 - 2 = {a}^{2} + \frac{1}{{a}^{2} }}}$

${\rm{7 = {a}^{2} + \frac{1}{ {a}^{2} } }}$

Answer is 7.

★ Formula used:

${\rm{\red{\boxed{(a+b)² = a²+2ab+b²}}}}$

Answer 2

EVALUATION ★

GIVEN★

$= > a + \frac{1}{a} = 3 ...(1$

find★

$= a {}^{2} + \frac{1}{a {}^{2} }$

solution★

• EQ (1 both side doing squaring

$= > (a + \frac{1}{a} ) {}^{2} = 3 {}^{2}$

we know that

$= > (a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab$

so

$= > a {}^{2} + \frac{1 {}^{} }{a {}^{2} } + 2(a)( \frac{1}{a} ) = 9 \\ \\ = > a {}^{2} + \frac{1}{a {}^{2} } + 2 = 9 \\ \\ = > a {}^{2} + \frac{1}{a {}^{2} } = 9 - 2 \\ \\ = > a {}^{2} + \frac{1}{a {}^{2} } = 7$

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answer★:

option (c) is ✅✅

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I hope it helps you