Question

If a = √5 - 2, then a -1/a is _______ *​

Answer 1

Step-by-step explanation:

Given:-

a = √5-2

To find:-

Find the value of a -(1/a) ?

Solution:-

Given that

a = √5 -2

1/ a = 1/(√5-2)

We know that

The Rationalising factor of √a-b = √a+b

The Rationalising factor of √5-2 = √5+2

On Rationalising the denominator then

=>[ 1/(√5-2) ]×[ (√5+2)/(√5+2)]

=> (√5+2)/[(√5-2)(√5+2)]

The denominator is in the form of (a+b)(a-b)

Where a = √5 and b=2

We know that

(a+b)(a-b) = a^2-b^2

=>(√5+2)/[(√5)^2-2^2]

=> (√5+2)/(5-4)

=>(√5+2)/1

=>√5+2

1/a = √5 +2

Now

a -(1/a)

=> (√5-2)- (√5+2)

=>√5-2 - √5 -2

=>(√5-√5)+(-2-2)

=>-4

Answer:-

The value of a-(1/a) for the given problem is -4

Used formulae:-

• The Rationalising factor of √a-b = √a+b

• (a+b)(a-b) = a^2-b^2

Answer 2

Answer:

$\large \blue{ \fcolorbox{purple}{lavenderblush}{ \: \textsf{\ \ \ \ddag \ \ \ \pink{QUESTION}\ \ \ \ddag \ \ \ }}}$

a = √5 - 2

$\frac{ 1}{a} = \frac{1}{ \sqrt{5} - 2} \\ \\ \sf{\implies = \frac{1}{ \sqrt{5} - 2} \times \frac{ \sqrt{5} + 2 }{ \sqrt{5} + 2}} \\ \\ \sf{\implies\frac{1( \sqrt{5} + 2) }{ ({ \sqrt{5} })^{2} - ({2})^{2}} } \\ \\ \sf{\implies \frac{ \sqrt{5} + 2}{5 - 4}} \\ \\ \sf{\implies \frac{ \sqrt{5} + 2}{1} } \\ \\ \sf{\implies \sqrt{5} + 2}$

So, value of $\frac{1}{a}$ = √5 + 2

$a - \frac{1}{a} = ( \sqrt{5} - 2) - ( \sqrt{5} + 2) \\ \sqrt{5} - 2 - \sqrt{5} - 2 \\ - 2 - 2 \\ \boxed{ - 4}$

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