Question

If p=5-3√3, then find the value of p+1/p. Also state that whether it is rational or irrational.
Please explain with steps and I shall mark you Brainliest answer​

Answer 1

Given :-

p = 5 - 3√3

Required to find :-

• The value of p + 1/p ?

• Whether the value is irrational number ot not ?

Solution :-

Given data :-

p = 5 - 3√3

Let's find the value of p + 1/p

=> p = 5 - 3√3

=>1/p = 1/5 - 3√3

Let's rationalise the denominator

Rationalising factor of 5 - 3√3 is 5 + 3√3

Multiply the numerator and denominator with the rationalising factors .

$\tt\dfrac{1}{5 - 3 \sqrt{3} } \times \dfrac{5 + 3 \sqrt{3} }{5 + 3 \sqrt{3} }$

$\tt \dfrac{5 + 3 \sqrt{3} }{(5 - 3 \sqrt{3})(5 + 3 \sqrt{3}) }$

Here we need to use an algebraic identity ;

i.e.

• ( x + y ) ( x - y ) = x² - y²

$\tt \dfrac{5 + 3 \sqrt{3} }{(5 {)}^{2} - ( 3 \sqrt{3} {)}^{2} }$

$\tt \dfrac{5 + 3 \sqrt{3} }{5 \times 5 - 3 \times 3 \times \sqrt{ {3}^{2} } }$

$\tt \dfrac{5 + 3 \sqrt{3} }{25 - 9 \times 3}$

$\tt \dfrac{5 + 3 \sqrt{3} }{25 - 27}$

$\tt \dfrac{5 + 3 \sqrt{3} }{ - 2}$

$\tt value \: of \: \dfrac{1}{p} = \bf \dfrac{5 + 3 \sqrt{3} }{ - 2}$

This implies ;

$\tt p + \dfrac{1}{p} \\ \\ \longrightarrow \tt 5 - 3 \sqrt{3} + \dfrac{5 + 3 \sqrt{3} }{ - 2} \\ \\ \longrightarrow \tt \dfrac{(5 - 3 \sqrt{3}) - 2 }{1 \times - 2} + \dfrac{5 + 3 \sqrt{3} }{ - 2} \\ \\ \longrightarrow \tt \dfrac{ - 10 + 6 \sqrt{3} }{ - 2} + \dfrac{5 + 3 \sqrt{3} }{ - 2} \\ \\ \implies \tt \dfrac{ - 10 + 6 \sqrt{3} + 5 + 3\sqrt{3} }{ - 2} \\ \\ \implies \tt \dfrac{ - 5 + 9 \sqrt{3} }{ - 2}$

Hence,

$\sf{Value \: of \: p + \dfrac{1}{p} = \dfrac{ - 5 + 9 \sqrt{3} }{ - 2} }$

similarly,

$\tt{ \dfrac{ - 5 + 9 \sqrt{3} }{ - 2}} \: \: is \: \: an \: \: irrational \: number \:$

Additional Info :-

Now,

Let's prove that the -5+9√3/-2 is a irrational number

Let's assume on the contradictory that - 5 + 9√3/-2 is a rational number .

So,

Equal this with p/q ( where p , q are integers , q ≠ 0 , p and q are co - primes )

This implies ;

$\tt \dfrac{ - 5 + 9 \sqrt{3} }{ - 2} = \dfrac{p}{q}$

$\tt - 5 + 9 \sqrt{3} = \dfrac{p}{q} \times - 2$

$\tt - 5 + 9 \sqrt{3} = \dfrac{ - 2p}{q}$

$\tt 9\sqrt{3} = \dfrac{ - 2p}{q} + 5$

$\tt 9 \sqrt{3} = \dfrac{ - 2p + 5q}{q}$

$\tt \sqrt{3} = \dfrac{ - 2p + 5q}{ \dfrac{q}{9} }$

$\tt \sqrt{3} = \dfrac{ - 2p + 5q}{9q}$

we know that ;

√3 is an irrational number

But,

$\tt\dfrac{-2p+5q}{9q}$ is a rational number .

This implies ;

$\tt \sqrt{3} \neq \dfrac{ -2p + 5q}{9q}$

Because,

An irrational number is never equal to a rational number .

Hence proved ! ✅

Answer 2

P=5-3√3

p+1/p

=5-3√3+1/5-3√3

=5-3√3+1/5-3√3×5+3√3/5+3√3

5-√3+5+√3/25-9×3

5-√3+5+√3/-2

-10+6√3+5+√3/-2

-5+9√3/-2

And it is irrational number