Question

(2+√2)²rational or irrational​

Answer 1

$\huge\boxed{\underline{\bf { \red S \green o \pink L \blue u \orange T \purple i\red O \pink n \green{..}}}}$

Step-by-step explanation:

$Given, {( \sqrt{2} - 2)}^{2}$

To find

• Rational or irrational

$\sf\mathbb\color{red} {solution}$

$\sf\mathbb\color{blue} { given \: = ( {( \sqrt{2} - 2) }^{2} }$

$\sf\mathbb\color{blue} { = {( \sqrt{2}) }^{2} - 2 \times \sqrt{2} \times 2 + {2}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\mathbb\color{yellow} { [By \: algebraic \: identity, }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\mathbb\color{yellow} { {(x - y) }^{2} = {x}^{2} - 2xy + {y}^{2}] }$

$\sf\mathbb\color{blue} { = 2 - 4 \sqrt{2} + 4 }$

$\sf\mathbb\color{blue} { = 6 - 4 \sqrt{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\mathbb\color{pink} {[irrational \: number \: ] }$

Therefore,

$\: \: \sf\mathbb\color{purple} { {( \sqrt{2} - 2) }^{2} is \: an \: irrational \: number \: }$

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Formula used

$\sf\mathbb\color{red} { {( {a} - b)}^{2} = {a}^{2} - 2ab + {b}^{2} }$