Question

Prove that 3+2root7 us an irrational number

Answer 1

$\huge{\underline{\mathfrak{\textcolor{blue}{Answer :}}}}$
$\huge\boxed{\fcolorbox{red}{pink}{irrational}}$
$\huge{\underline{\mathrm{\textcolor{red}{Step-by-step \: \: explanation :}}}}$

$\LARGE{\underline{\mathtt{\textcolor{violet}{Given :-}}}}$
⚪ [ 3 + 2√7 ]

$\LARGE{\underline{\mathtt{\textcolor{green}{To \: \: prove :-}}}}$
⚪ $3 + 2 \sqrt{7}$ is an irrational number.

$\LARGE{\underline{\mathtt{\textcolor{teal}{Concept \: \: used :-}}}}$
⚪ Real numbers
⚪ Irrational numbers

$\LARGE{\underline{\mathtt{\textcolor{blue}{Proof :-}}}}$
✒ If possible, $3 + 2 \sqrt{7}$ be a rational number.

let $3 + 2 \sqrt{7} = \Large{ \frac{a}{b} }$, where a and b are co-primes and b ≠ 0.

Then,
✒ $3 + 2 \sqrt{7} = \Large{ \frac{a}{b} }$

On squaring both the sides :

$\Rightarrow { \left(3 + 2 \sqrt{7} \right) }^{2} = \Large{ { \left( \frac{a}{b} \right) }^{2} }$

$\Rightarrow 9 + 28 + 12 \sqrt{7} = \Large{ \frac{ {a}^{2} }{ {b}^{2} } }$

$\rightarrow 37 + 12 \sqrt{7} = \frac{ {a}^{2} }{ {b}^{2} } \\ \\ \rightarrow \sqrt{7} = \frac{ {a}^{2} - 37 {b}^{2} }{12 {b}^{2} }$
Here, a and b are integers.

Therefore, $\sf \: \frac{ {a}^{2} - 37 {b}^{2} }{12 {b}^{2} } \: is \: a \: rational \: number.$
Thus, $\sf \: \sqrt{7} \: is \: also \: a \: rational \: number.$

But, this contradicts the fact that $\sqrt{7}$ is an irrational number.
This contradiction arises on assuming $3 + 2 \sqrt{7}$ to be a rational number.
So, our assumption is wrong.
Hence, $3 + 2 \sqrt{7}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$.

$\LARGE{\underline{\mathtt{\textcolor{magenta}{Conclusion :-}}}}$
⚪$3 + 2 \sqrt{7}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$.

Kindly give thanks to my this answer : $<a href="https://brainly.in/question/20136092" alink="green" vlink="red">THANKS</a>$
.
.
.
.
.
.
.
$\bold{hope………. \: this \: answer \: helps \: you \: the \: best .}$
and…… mark this answer as ❤BRAINLIEST❤
and. ….. If you can… .then

Thanks for reading…………………………………………………….. =)
$\huge{\fcolorbox{orange}{orange}{brainliest answer}}$